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2007-12-31T20:49:00.001-08:00

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:::|Sweet Jokes |::: HAPPY NEW YEAR

2007-12-31T20:49:00.000-08:00

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:::|Sweet Jokes |::: Happy New Year 2008

2007-12-31T20:43:00.000-08:00

Wishing You all a very Happy and Prosperous New Year 2008

From
Anurag Bhatia
Sweet Jokes

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:::|Sweet Jokes |::: Happy New Year 2008

2007-12-31T20:42:00.000-08:00

I Wish you

colorful

HAPPY NEW YEAR

2008

May god bless you and be with you in all the way of your life.

May this New Year brings all your dreams come true

Though New Year comes only once a year.....

My wishes for you and your family is continuing

LOVE, PEACE & HAPPINESS

Throughout the year. .

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:::|Sweet Jokes |::: Happy New Year 2008..........

2007-12-31T19:59:00.000-08:00

ope for the best and never forget that anything is possible as long as you remain dedicated to the task.

ccept others for who they are and for the choices they've made even if you have difficulty understanding their beliefs, motives, or actions.

lay;Never forget to have fun along the way. Success means nothing without happiness.

ield to commitment. If you stay on track and remain dedicated, you'll find success at the end of the road.

ever ignore the poor, infirm, helpless, weak, or suffering. Offer your assistance when possible, and always your kindness and understanding.

xplore and experiment. The world has much to offer, and you have much to give. And every time you try something new, you'll learn more about yourself.

ork hard every day to be the best person you can be, but never feel guilty if you fall short of your goals. Every sunrise offers a second chance.

ield to commitment. If you stay on track and remain dedicated, you'll find success at the end of the road.

xplore and experiment. The world has much to offer, and you have much to give. And every time you try something new, you'll learn more about yourself.

ccept others for who they are and for the choices they've made even if you have difficulty understanding their beliefs, motives, or actions.

efuse to let worry and stress rule your life, and remember that things always have a way of working out in the end.

2008

From:

Deepak Kumar Gupta

HARIDWAR

Mb: +919837166366

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:::|Sweet Jokes |::: New Year Wishes

2007-12-31T19:42:00.000-08:00

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Re: [Math4u] Re: i am in desperate need of help with math

2007-12-31T13:04:00.000-08:00

yeah thanks. that kind of stuff is easy ;) its the more in depth stuff..

----- Original Message ----
From: Kirk Bevins <hatton02@yahoo.co.uk>
To: Math4u@yahoogroups.com
Sent: Monday, December 31, 2007 1:46:28 PM
Subject: [Math4u] Re: i am in desperate need of help with math

--- In Math4u@yahoogroups. com, "lizjones2008" <lizjones2008@ ...> wrote:
>
> Hi. I have taken this math course at school (cecil college) twice. I
am
> clueless when it comes to anything other than basic math. This math
> course is algebra and trig. I am in major need of help. Is there
anyone
> who is interested in tutoring and or helping me? Please and thank
you!
> -L
>

Algebra IS basic maths ;). Oh, and I'm in England so I can't help you.
I haven't got the energy for online tutoring any more. Algebra is just
arithmetic hidden in numbers.

3a means 3 x a where 'a' is any number. If a was equal to 6, then 3a =
3 x 6 = 18. Learning more rules like this, it becomes easy.

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[Math4u] Re: A combinations/probability problem

2007-12-31T12:40:00.001-08:00

There are only a few rules used for computing likeliness. You want
to pick 6 numbers out of 20 available. So for the first pick you
have 20 equally likely outcomes. Let's say you want to pick them in
the order given. You have 2 desirable outcomes. So the likeliness
of getting a desirable outcome on the first pick is (2/20). Next you
want to pick a 3. There are 2 threes and 19 remaining numbers. So
the likeliness of a desirable second pick is (2/19).

The likeliness of getting 6 desirable picks in a row (1, 3, 4, 5, 6,
7 ) is (likeliness 1)*(likeliness 2)*(likeliness 3)* (likeliness 4)*
(likeliness 5)*(likeliness 6)
Or (2/20)*(2/19)*(2/18)*(2/17)*(2/16)*(2/15)
=2^6/(20*19*18*17*16*15)

The likeliness of picking 1, 1, 2, 2, 3, 3 is computed as follows:
Pick first 1, 2 available out of 20 total = 2/20
Pick second 1, 1 available out of 19 total = 1/19
Pick first 2, 2 available out of 18 total = 2/18
Pick second 2, 1 available out of 17 total = 1/17
Etceteras
Or (2/20)*(1/19)*(2/18)*(1/17)*(2/16)*(1/15)
=2^3/(20*19*18*17*16*15)

The likeliness of first ;pick in any order:
(12/20)*(10/19)*(8/18)*(6/17)*(4/16)*(2/15)

The likeliness of the second pick in any order:
(6/20)*(5/19)*(4/18)*(3/17)*(2/16)*(1/15)

As a check,

the likeliness of the second pick in order is 2^8 less than the first
pick in order because in 3 of the picks, there is one available
remaining instead of 2.

the likeliness of the first pick in any order is 6! times the first
pick in the specified order. There are 6! ways of arranging the
order because you have six to choose from for the first choice, and
one fewer choices for each choice that follows.

Regards,
Brian

--- In Math4u@yahoogroups.com, "Kirk Bevins" <hatton02@...> wrote:
>
> OK. Imagine you are selecting 6 numbers from a random selection of
20
> numbers. The numbers are:
>
> 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10.
>
> If anybody has seen the British TV show, Countdown, then this is
> where the problem comes from.
>
> How much more likely is it of choosing 1, 3, 4, 5, 6, 7 compared
with
> 1, 1, 2, 2, 3, 3?
>
> The next question (which probably helps solve the above) is:
> There are 6! ways of arranging 1, 3, 4, 5, 6, 7 in some order. How
> many ways are there of arranging 1, 1, 2, 2, 3, 3 in some order?
This
> kind of thing I struggle with as we have repeated numbers. Do we
just
> do 6! divided by 3 as there are 3 repeats? I don't have the right
> intuition for combinations.
>
> Any explanation as to how you worked these out would be
appreciated.
> I don't understand combinations so don't feel like you're
patronising
> if you explain fully!!
>
> Kirk
>

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:::|Sweet Jokes |::: A billion dollars

2007-12-31T11:46:00.000-08:00

A Veteran - whether active duty, retired, guard, or reserve -
is someone who, at some point in his life, wrote a check made
payable to "The United States of America" for an amount of
"up to and including my life".
That is Honor,
and there are too many people
in this country
who no longer understand it ! ! ! !

No, the United States of America is by no means a perfect country,
but it's my country and I shall defend her 'til the day I die!

***********************************************************************

A billion

This is too true to be funny

The next time you hear a politician use the

word "billion" in a casual manner, think about

whether you want the "politicians" spending

YOUR tax money.

A billion is a difficult number to comprehend,

but one advertising agency did a good job of

putting that figure into some perspective in

one of its releases.

A. A billion seconds ago it was 1959.

B. A billion minutes ago Jesus was alive.

C. A billion hours ago our ancestors were

living in the Stone Age.

D. A billion days ago no-one walked on the earth on two feet.

E. A billion dollars ago was only 8 hours and

20 minutes, at the rate our government is spending it.

While this thought is still fresh in our brain, let's take a look at New Orleans It's amazing what you can learn with some simple division . .

Louisiana Senator, Mary Landrieu (D), is presently asking the Congress for $250 BILLION to rebuild New Orleans. Interesting number, what does it mean?

A. Well, if you are one of 484,674 residents of

New Orleans (every man, woman, child), you </ FONT>

each get $516,528.

B. Or, if you have one of the 188,251 homes in

New Orleans , your home gets $1,329,787.

C. Or, if you are a family of four, your family

gets $2,066,012.

Washington, D.C .. HELLO!!! ... Are all your calculators broken??

Inheritance TAX!!

Accounts Receivable Tax
Building Permit Tax
CDL License Tax
Cigarette Tax
Corporate Income Tax
Dog License Tax
Federal Income Tax
Federal Unemployment Tax (FUTA) < BR>Fishing License Tax
Food License Tax
Fuel Permit Tax
Gasoline Tax
Hunting License Tax
Inheritance Tax
Inventory Tax
IRS Interest Charges (tax on top of tax),
IRS Penalties (tax on top of tax),
Liquor Tax,
Luxury Tax,
Marriage License Tax,
Medicare Tax,
Property Tax,
Real Estate Tax,
Service charge taxes,
Social Security Tax,
Road Usage Tax (Truckers),
Sales Taxes,
Recreational Vehicle Tax,
School Tax,
State Income Tax,
State Unemployment Tax (SUTA),
Telephone Federal Excise Tax,
Telephone Federal Universal Service Fee Tax,
Telephone Federal, State and Local Surcharge Tax,
Telephone Minimum Usage Surcharge Tax,
Telephone Recurring and Non-recurring Charges Tax,
Telephone State and Local Tax,
Telephone Usage Charge Tax,
Utility Tax,
Vehicle License Registration Tax,
Vehicle Sales Tax,
Watercraft Registration Tax,
Well Permit Tax,
Workers Compensation Tax.

STILL THINK THIS IS FUNNY?
Not one of these taxes existed 100 years ago,
and our nation was the most prosperous in the world.
We had absolutely no national debt, had the largest middle class in the world, and Mom stayed home to raise the kids.

What happened? Can you spell 'politicians!'

And I still have to "press

1" for English.

I hope this goes around THE
USA at least 100 times

What the heck happened?????

__,_._,___

Re: [Math4u] A combinations/probability problem

2007-12-31T10:49:00.000-08:00

So, ordering is important in your problem?

any way, I guess it can be solved with ordering or without ordering. The answer is the same.

There is a formula for arrangements with repeatition Kirk.

As for arranging 1,1,2,2,3,3:

6! / (2! * 2! * 2!), that is 6!/8 which is 6*5*3 = 90 ways.

Kirk Bevins <hatton02@yahoo.co.uk> wrote:

OK. Imagine you are selecting 6 numbers from a random selection of 20
numbers. The numbers are:

1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10.

If anybody has seen the British TV show, Countdown, then this is
where the problem comes from.

How much more likely is it of choosing 1, 3, 4, 5, 6, 7 compared with
1, 1, 2, 2, 3, 3?

The next question (which probably helps solve the above) is:
There are 6! ways of arranging 1, 3, 4, 5, 6, 7 in some order. How
many ways are there of arranging 1, 1, 2, 2, 3, 3 in some order? This
kind of thing I struggle with as we have repeated numbers. Do we just
do 6! divided by 3 as there are 3 repeats? I don't have the right
intuition for combinations.

Any explanation as to how you worked these out would be appreciated.
I don't understand combinations so don't feel like you're patronising
if you explain fully!!

Kirk

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[Math4u] Re: i am in desperate need of help with math

2007-12-31T10:46:00.001-08:00

Algebra IS basic maths ;). Oh, and I'm in England so I can't help you.
I haven't got the energy for online tutoring any more. Algebra is just
arithmetic hidden in numbers.

3a means 3 x a where 'a' is any number. If a was equal to 6, then 3a =
3 x 6 = 18. Learning more rules like this, it becomes easy.

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Re: [Math4u] Another problem about balance

2007-12-31T10:36:00.000-08:00

an old one which shows the power of powers.

minimum is 4 weights: 3^0, 3^1, 3^2, and 3^3.

Mosaad

renato ab <renato_eq@yahoo.com.br> wrote:

I think you'll get it easily.

You have a balance of pans and you need to weigh every integer weight from 1 kg to 40 kg. What's the minimum number of known weights you need to use? What are their mass?

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:::|Sweet Jokes |::: Happy & Prosperous 2008

2007-12-31T09:06:00.000-08:00

May this new year bring you and your family loads of success, good health, happiness & joy!!

Wish You all a Very Happy & Prsoperous New Year !

Regards,

Sujin

__,_._,___

[Math4u] Re: Questions ???

2007-12-31T07:52:00.001-08:00

Hi everyone can you help me to solve these questions:
1.Calculate [ ( -2 ) ^ 0 y ^ -1 ( 2y ) ^ 3 / ( 2y ^ -2 ) ^ -1 y ^ -
4 ]
2. Use the absolute value notation " | | " to describe :
a. x is farther from 0 than it is from 5 .
b. x is more than b units from a but less than c units from a .

Brian's solution:
The first problem is not properly written and intended to give the
student grief.
Look at the part 0y^
Looking at it from this view, it is firmly agreed to mean …0*y^ …
Looking at the view (-2)^0y, it can be argued either way whether it
means 0*y or (0*y). However the second exponential sign means we
must accept the first interpretation.

However, from what I've seen, there are only certain times when
formulas can be written without using the multiplication sign.

In the formula (2x+1)(x+3)=2x^2+7x+3 there 4 multiplication signs
that are understood but not written. We can put bracketed terms or
exponentials together or other simple terms such as trig functions
together and it is understood that they are multiplied. We can put a
number in front of terms and it is understood that they are
multiplied. It is understood because it is done very commonly. It
is introduced to students without thinking and the students pick it
up automatically. If some stupid teacher tried to explain it, the
students would never grasp it.

However, if you have (-2)^0 times y^-1, then you cannot put them
together without a multiplication sign. This is never done. You
could if the 0 was written on a higher level as a superscript.

Mathematics doesn't have a lot of rules. Some you will not find this
written anywhere. It is like working for a company. There are too
many unwritten rules to write down, but if you violate any of them,
you will be on the way out and everyone else will know it.

When I see a term such as 0y or –1y or –1(2y), I think of the term as
being something that cannot be broken apart. That is why this
equation will throw people off.

However, it we have
0y^ always by convention means 0*y^
-1(2y)^ always by convention means –1*(2y)^
-1y^-4 always by convention means –1*y^-4

Your equation should never have been written in such a confusing and
incorrect manner, but after thinking about it for a couple days, the
only possible way to evaluate it is as:
[ ( -2 ) ^ 0 * y ^ -1 * ( 2y ) ^ 3 / ( 2y ^ -2 ) ^ -1* y ^ -4 ]
It then becomes trivial:
( -2 ) ^ 0 * y ^ -1 * ( 2y ) ^ 3 / ( 2y ^ -2 ) ^ -1* y ^ -4
= + + +1 * y ^ -1 * ( 2y ) ^ 3 / ( 2y ^ -2 ) ^ -1* y ^ -4
= + + + + y ^ -1 * ( 2y ) ^ 3 / ( 2y ^ -2 ) ^ -1* y ^ -4
= + + + + + + + + 8 * y ^ 2 / ( 2y ^ -2 ) ^ -1* y ^ -4
Here's another trap. Many students think the denominator after the
slash extends to the right forever, but it doesn't.
= + + + + + + + + 8 * y ^ 2 * ( 2y ^ -2 ) * y ^ -4
= + + + + + + + + 16* y ^ -4
= 16* y ^ -4 answer
I'm guessing your teacher wanted the answer to be 16. However when
we divide by (2^-2)^-1, this is like a triple negative. I stand by
my answer.

Question 2 is more fun. I agree with Rich's answer, but would carry
it further
2. Use the absolute value notation " | | " to describe :
a. x is farther from 0 than it is from 5 .
b. x is more than b units from a but less than c units from a .

Solution:
By farther, they mean absolute value of the difference between x and
0
which can be written
Ix-0I or I0-xI or IxI. You can use the greater than sign < or > to
indicate that the distance is greater.
Ix-0I>Ix-5I
Now this wasn't asked for, but we can square both sides. Since both
sides are positive, the direction of the greater than sign will not
change.
x^2 - 0x +0 > x^2 – 10x + 25
We can subtract x^2 from both sides.
0 >– 10x + 25
We can add 10x to both sides
10x > 25
We can divide both sides by 10
x>2.5
Regards,
Brian

--- In Math4u@yahoogroups.com, dghbosa alias <redgate3000@...> wrote:
>
> Hi everyone can you help me to solve these questions :
>
> 1.Calculate [ ( -2 ) ^ 0 y ^ -1 ( 2y ) ^ 3 / ( 2y ^ -2 ) ^ -1 y
^ -4 ]
>
> 2. Use the absolute value notation " | | " to discribe :
>
> a. x is farther from 0 than it is from 5 .
>
> b. x is more than b units from a but less than c units from a .
>
>
> ---------------------------------
> Be a better friend, newshound, and know-it-all with Yahoo! Mobile.
Try it now.
>

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[Math4u] i am in desperate need of help with math

2007-12-31T07:36:00.000-08:00

Hi. I have taken this math course at school (cecil college) twice. I am
clueless when it comes to anything other than basic math. This math
course is algebra and trig. I am in major need of help. Is there anyone
who is interested in tutoring and or helping me? Please and thank you!
-L

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:::|Sweet Jokes |::: WISH U HAPPY NEW YEAR 2008....

2007-12-31T07:16:00.000-08:00

Hi,

Wish U and Ur family,

A Prosperous New Year

filled with memorable moments!!

Thanks & Regards,

s.prabhu

electrical engineer

BAHRAIN.

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[Inside AdSense] A look back at 2007

2007-12-31T07:07:00.000-08:00

2007 is drawing to a close, but before you get out the party hats and start practicing the lyrics to Auld Lang Syne, we'd like to take a moment to recap some of the biggest developments in AdSense this year.

New feature launches were abundant in 2007, and many were based on your direct feedback and requests. To help improve the performance of your ads, we revamped the look of our ad formats and introduced ad placements to let you promote your site to potential advertisers. The new Manage Ads feature gave you the ability to make changes to your ad units directly in your account and quickly experiment with optimizations. In addition, we launched such features as video units, AdSense for mobile content, and referrals 2.0 to bring in new revenue streams for your site. Still others like the Allowed Sites list and the recently announced Ad Review Center were inspired by your requests for more control over your ads and your account.

With our growing community of publishers around the world, we also continued our international expansion of products and online resources. AdSense for content was launched in languages such as Hebrew, Greek, and Bulgarian, while AdSense for search became available in a number of Southeast Asian languages. We added payment options such as Western Union Quick Cash in a few countries and EFT in Mexico. To help publishers connect with one another around the world, we introduced AdSense Help Forums in 9 new languages, including Polish, Arabic, and Russian.

And of course, we can't forget to mention the exciting things that happened right here on our blog. We welcomed 6 new languages into the Inside AdSense family (Italian, Korean, Japanese, French, Turkish, and Simplified Chinese), bringing the total number of AdSense blogs to 11. All 11 blogs combined received 2.6 million pageviews from 1.7 million visits, and global Groups subscriptions to all blogs crossed the 43,000 mark.

Finally, we'd like to close with our second annual Reader's Choice Award, which goes to the post you visited most this year: 'Introducing video units'. Accepting the award on behalf of the post is author Ryan Hayward, who had this to say: "In a world where Hollywood writers are striking and my favorite show, 'The Office,' is off-air, I hope that announcing video units helped to fill the creative void and made each of our lives just a little more entertaining. Cue the orchestra."

Here's to a very happy and prosperous 2008. Cheers!

Posted by Arlene Lee - AdSense Publisher Support

--
Posted By Inside AdSense Team to Inside AdSense at 12/31/2007 07:07:00 A
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:::|Sweet Jokes |::: Happy New Year 2008 !!!!!!!!!!!!!!!!!!!!!!!!!!

2007-12-31T05:48:00.000-08:00

Dawning of a New Year - New Hopes New Joys and New Beginnings.

Here's Wishing each new day brings for you Happiness ,
Prosperity and continued success!!!!!!

************** HAPPY NEW YEAR 2008 *****************

NITIN SHARMA

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[forwards4all] -:|:- Beauty of Snow -:|:-

2007-12-31T01:47:00.000-08:00

Link of the Day: Dual Notebook

New Year SMS

neeli.ankhain@yahoo.co.in

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[Math4u] Leaving Vegas

2007-12-31T01:32:00.000-08:00

Before you go playing games of chance involving dice, try our online probability program that allows you to compute probabilities for upto 3 dice (whether fair or unfair). It will report on probabilities of individual dice, joint or compound probability of upto 3 dice, and probabilities for sum of digits.

Visit http://www.thinkanddone.com/ge/Dice.aspx

Asad

__._,_.___

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[forwards4all] Creative Ads !!! Funny Enough :)

2007-12-31T00:35:00.000-08:00

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[forwards4all] New Year Wishes 4 U !!!

2007-12-31T00:31:00.000-08:00

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[forwards4all] PENALTY CORNER FOR EVERY MISTAKE !!!

2007-12-30T23:59:00.000-08:00

[forwards4all] Happy New Year - 2008

2007-12-30T23:16:00.000-08:00

[forwards4all] !!~ New Year Greetings For My Angel friends~!!

2007-12-30T23:04:00.000-08:00

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[forwards4all] !!~ Cute Cute New Ye@r Gifs ~!!

2007-12-30T22:29:00.000-08:00

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[forwards4all] +++ Happy New Year 2008+++

2007-12-30T20:58:00.000-08:00

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[Math4u] What it could be named?

2007-12-30T19:58:00.000-08:00

Hi
12692 digits in one figure. Is any name for it.?
Thanks

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[Math4u] A combinations/probability problem

2007-12-30T17:39:00.001-08:00

OK. Imagine you are selecting 6 numbers from a random selection of 20
numbers. The numbers are:

1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10.

If anybody has seen the British TV show, Countdown, then this is
where the problem comes from.

How much more likely is it of choosing 1, 3, 4, 5, 6, 7 compared with
1, 1, 2, 2, 3, 3?

The next question (which probably helps solve the above) is:
There are 6! ways of arranging 1, 3, 4, 5, 6, 7 in some order. How
many ways are there of arranging 1, 1, 2, 2, 3, 3 in some order? This
kind of thing I struggle with as we have repeated numbers. Do we just
do 6! divided by 3 as there are 3 repeats? I don't have the right
intuition for combinations.

Any explanation as to how you worked these out would be appreciated.
I don't understand combinations so don't feel like you're patronising
if you explain fully!!

Kirk

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[forwards4all] In The New Year

2007-12-30T09:54:00.000-08:00

WOH BHEEGEY PAL

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2008

In The New Year

In the New Year,
we wish you the best year you've ever had,
and that each New Year
will be better than the last.
May you realize your fondest dreams
and take time to recognize and enjoy
each and every blessing.

Happy New Year, And many more!

God bless your New Year
With His presence ever near

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:::|Sweet Jokes |::: File - Join Sweet Jokes

2007-12-30T07:51:00.000-08:00

Hello Everybody!

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[forwards4all] True Friendship

2007-12-30T07:09:00.000-08:00

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[forwards4all] The Best Pictures of 2007

2007-12-30T07:08:00.000-08:00

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[Math4u] Another problem about balance

2007-12-30T05:30:00.000-08:00

I think you'll get it easily.

You have a balance of pans and you need to weigh every integer weight from 1 kg to 40 kg. What's the minimum number of known weights you need to use? What are their mass?

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[forwards4all] New Speed Limit Device

2007-12-30T05:25:00.000-08:00

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Re: [Math4u] Re: Cube problem

2007-12-30T04:43:00.001-08:00

Hi,
Great effort. My answers are:-
Answer 1: 15.
Answer 2: 45.
Answer 3: 30.
Sanjiva Dayal

--- Brian Edward Jensen <brianejensen@prodigy.net>
wrote:

> OK,
> In that case (where for [1,2,3,4] 1 and 2 are
> opposite, 3 and 4 are
> opposite) I believe
> [1234] and [1243] and [3412] are duplicates
> [1235] and [1253] and [3512] are duplicates
> [1236] and [1263] and [3612] are duplicates
> For example we can rotate [1,2,3,4] about the 1,2
> axis and get
> [1,2,4,3]
> 135 printed out ways divided by 3 duplicates in each
> group equals 45,
> ANSWER.
>
> Another way of looking at the problems:
> Question 1, 6 different colors; find the number of
> ways of picking
> unique colors for 1 pair of opposite sides.
> Question 2, 6 different colors; find the number of
> ways of picking
> unique colors for 2 pairs of opposite sides.
> Question 3, 6 different colors; find the number of
> ways of picking
> unique colors for 3 pairs of opposite sides.
>
> I am feeling more confident in my answers
> Question 1 we agree the answer is 6*5/2=15
> Question 2
> Ways of picking 4 colors, any order =
> 6*5*4*3/(4*3*2*1)=15
> Ways of excluding 2 colors, any order = 6*5/(2*1)=15
> Pick any color. It has 3 choices for the opposite
> color. So the
> answer is
> 15*3=45
> Question 3
> Method A
> Pick any color.
> You have 5 choices for the opposite side.
> Pick any of the remaining 4 colors.
> You have 3 choices for the opposite side
> You have 2 choices for the next 2 colors.
> 5*3*2=30 Answer
> Method B
> Assign a color to a side.
> You have 5*4*3*2*1 or 120 ways of assigning the
> other colors.
> Keeping our chosen color to the front, we can rotate
> the cube into 4
> positions.
> 120/4=30 Answer
> Method C
> Don't move the cube
> Ways of assigning the colors = 6*5*4*3*2*1=720
> There are 6 different faces that can be put in front
> and for each
> face the cube can be rotated into 4 positions so
> there are 24
> orientations of the cube.
> 720/24=30 answer.
> Method D
> We take our answer of 45 from question 2. There are
> 2 ways to add
> the last two colors. But we will have three
> duplicates for each
> combination.
> 45*2/3=30 answer.
> Regards,
> Brian
>
> --- In Math4u@yahoogroups.com, "w7anf" <cherry@...>
> wrote:
> >
> > Sides are numbered 1 bottom, 2 top, 3 left, 4
> right, 5 back, 6
> front.
> > [1,2,3,4] means color 1 is on the bottom.
> > Color 2 is opposite on the top.
> > Color 3 is on the left.
> > Color 4 is opposite on the right.
> > [1,4,2,3] is different and can not be rotated to
> [1,2,3,4].
> > Jim FitzSimons
> >
> > --- In Math4u@yahoogroups.com, "Brian Edward
> Jensen"
> > <brianejensen@> wrote:
> > >
> > > Jim, you went to a lot of work!
> > > My answers are
> > > Question 1 = 15, you agree
> > > Question 2 = 45, Jim got 135
> > > Question 3 = 30
> > > I have a feeling I made another mistake. Seems
> to me that once
> you
> > > have the ring of 4 colors around the cube, there
> are only 2 ways
> > to
> > > assign the remaining 2 colors so answer 3 should
> be twice answer
> > 2.
> > > I'll look at it tonight.
> > > So Jim's answer is 3 times my answer for
> question 2. It could be
> > > interpretation. I look at Jim's list of 135
> possibilities. I
> would
> > > interpret [1234], [1432], and [3412] as
> duplicates because they
> > each
> > > have 1 and 3 opposite. If you divide 135 by 3
> you get 45. So Jim
> > and
> > > I are on the same track.
> > > Looking at Jim's solution for question 3, I
> can't tell which
> sides
> > > are opposite. If the three pairs of opposites
> were the same,
> there
> > > would still be two unique solutions. We could
> call them right
> hand
> > or
> > > left hand with the thumb on one axis pointing
> toward increasing
> > value
> > > and the fingers in the direction of rotation.
> Don't know how we'd
> > > decide the direction.
> > >
> > > Regards,
> > > Brian
> > >
> > > --- In Math4u@yahoogroups.com, "w7anf" <cherry@>
> wrote:
> > > >
> > > > Question:-
> > > > Given six different colours and a cube.
> > > > 1. In how many ways two opposite faces of the
> cube can
> > > > be coloured with two different colours?
> > > > 2. In how many ways four faces of the cube can
> be
> > > > coloured with four different colours of which
> two
> > > > faces are opposite and other two faces are
> also
> > > > opposite?
> > > > 3. In how many ways all six faces of the cube
> can be
> > > > coloured with six different colours?
> > > >
> > > > Colors 1,2,3,4,5,6
> > > > Sides 1 bottom, 2 top, 3 left, 4 right, 5
> back, 6 front
> > > >
> > > > 1. C(6,2)=6*5/2=15
> > > >
>
[[1,2],[1,3],[1,4],[1,5],[1,6],[2,3],[2,4],[2,5],[2,6],
> > > > [3,4],[3,5],[3,6],[4,5],[4,6],[5,6]]
> > > >
> > > > 2. 15*4*3=180
> > > > Flip and rotate.
> > > > Without duplicates there are
> > > > 135 different ways.
> > > >
> [[1,2,3,4],[1,2,3,5],[1,2,3,6],[1,2,4,3],[1,2,4,5],
> > > >
> [1,2,4,6],[1,2,5,3],[1,2,5,4],[1,2,5,6],[1,2,6,3],
> > > >
> [1,2,6,4],[1,2,6,5],[1,3,2,4],[1,3,2,5],[1,3,2,6],
> > > >
> [1,3,4,2],[1,3,4,5],[1,3,4,6],[1,3,5,2],[1,3,5,4],
> > > >
> [1,3,5,6],[1,3,6,2],[1,3,6,4],[1,3,6,5],[1,4,2,3],
> > > >
> [1,4,2,5],[1,4,2,6],[1,4,3,2],[1,4,3,5],[1,4,3,6],
> > > >
> [1,4,5,2],[1,4,5,3],[1,4,5,6],[1,4,6,2],[1,4,6,3],
> > > >
> [1,4,6,5],[1,5,2,3],[1,5,2,4],[1,5,2,6],[1,5,3,2],
> > > >
> [1,5,3,4],[1,5,3,6],[1,5,4,2],[1,5,4,3],[1,5,4,6],
> > > >
> [1,5,6,2],[1,5,6,3],[1,5,6,4],[1,6,2,3],[1,6,2,4],
> > > >
> [1,6,2,5],[1,6,3,2],[1,6,3,4],[1,6,3,5],[1,6,4,2],
> > > >
> [1,6,4,3],[1,6,4,5],[1,6,5,2],[1,6,5,3],[1,6,5,4],
> > > >
> [2,3,1,4],[2,3,1,5],[2,3,1,6],[2,3,4,5],[2,3,4,6],
> > > >
> [2,3,5,4],[2,3,5,6],[2,3,6,4],[2,3,6,5],[2,4,1,3],
> > > >
> [2,4,1,5],[2,4,1,6],[2,4,3,5],[2,4,3,6],[2,4,5,3],
> > > >
> [2,4,5,6],[2,4,6,3],[2,4,6,5],[2,5,1,3],[2,5,1,4],
> > > >
> [2,5,1,6],[2,5,3,4],[2,5,3,6],[2,5,4,3],[2,5,4,6],
>
=== message truncated ===

Sanjiva Dayal, B.Tech.(I.I.T. Kanpur)
Address:A-602, Twin Towers, Lakhanpur, Kanpur-208024, INDIA.
Phones:+91-512-2581532,2581426.
Mobile:9415134052
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Personal email:sanjivadayal@hotmail.com

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[forwards4all] Fwd: MICROWAVE

2007-12-30T01:53:00.000-08:00

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[forwards4all] Stall conversation...

2007-12-30T01:31:00.000-08:00

I was barely sitting down when I heard a voice from the other stall saying: "Hi, how are you?"

I'm not the type to start a conversation in the men's restroom but I don't know what got into me, so I answered, somewhat embarrassed, "Doin' just fine!"

And the other guy says: "So what are you up to?"

What kind of question is that? At that point, I'm thinking this is too bizarre so I say: "Uhhh, I'm like you, just travelling!"

At this point I am just trying to get out as fast as I can when I hear another question. "Can I come over?"

Ok, this question is just too weird for me but I figured I could just be polite and end the conversation. I tell him, "No........I'm a little busy right now!!!"

Then I hear the guy say nervously...
"Listen, I'll have to call you back. There's an idiot in the other stall who keeps answering all my questions !

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[forwards4all] Australia won 1st Test against India :(

2007-12-30T01:22:00.000-08:00

Regards,
¥@m!n!

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[forwards4all] COMMUNAL JOKES NOT ALLOWED

2007-12-30T00:05:00.000-08:00

Hello friends,

I'm receiving complaints from few members regarding jokes which are being circulated, targeted to a particular community. Its an inhuman act to make mockery of someone's culture, religion or ethnicity.

I request all the members not to post jokes which are targeted to particular religion or ethnic groups. Instead of that, you can always write a person's name as XYZ or Santa, Banta, etc.

Please note that COMMUNAL STUFF WILL NOT BE ENTERTAINED.

On behalf of the group, I apologize to our Sikh brothers if we have ever hurt their feelings and ensure that this will not be repeated.

Regards,

Moderator (forwards4all & hyd-masti groups)

----- Original Message ----

From: Mohinder Singh <mohindersingh_98@yahoo.com>
To: osman_hyderabad@yahoo.com
Sent: Saturday, 29 December, 2007 10:06:23 PM
Subject: COMMUNAL JOKES

dear Mr Osman

I am a member of the yahoo group of which I presume you are the moderator. I am writing to you to lodge a strong protest at the so-called "Sardarji jokes" which keep circulating in this yahoo group.

These jokes are directed at a particular community and are therefore nothing but COMMUNAL. I think this is quite obvious but I fail to see why you as a moderator have done nothing about them. This is all in poor taste and uncalled for.

what do u think will happen if tomorrow someone starts circulating jokes about other communities, say Hindu jokes or Muslim jokes? You can well imagine the problem it can create.

Please do something about it.

Thanks and regards,

Dr Mohinder Singh

Patiala

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[forwards4all] Life in pictures -Daughter of the east--BENAZIR BHUTTO

2007-12-29T23:19:00.000-08:00

WOH BHEEGEY PAL

http://groups.yahoo.com/group/woh_bheegey_pal/

Life in pictures -Daughter of the east

BENAZIR BHUTTO

Benazir Bhutto was part of a political dynasty. Here she is with India's foreign minister in 1972 in Simla, where her father, President Zulfikar Ali Bhutto, met India's Indira Gandhi.

Ms Bhutto in 1977, shortly after completing her education at Oxford University. Born in 1953 in the province of Sindh, she also attended Harvard.

Ms Bhutto was Pakistan's prime minister twice. The first term of office between 1988 and 1990 coincided with the latter years of UK PM Margaret Thatcher's rule.

During both her stints in power, the role of Ms Bhutto's husband, Asif Zardari, proved highly controversial as he was accused of stealing state funds - charges he denied.

Benazir Bhutto in 1991 during the rule of arch-rival Nawaz Sharif. She had been dismissed in 1990 on charges of corruption but was never tried.

Ms Bhutto was re-elected in 1993 and served as prime minister until 1996, again being dismissed for alleged corruption.

She remained in Pakistan until 1999 but then left to live abroad as questions remained about her and her husband's wealth

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[Math4u] Re: Questions ???

2007-12-29T20:53:00.001-08:00

Greetings:

If a and b each correspond to a point on the real line, then the distance from one point to the other is equal to |a - b| (or, equivalently, |b - a|). Thus the distance from x to 0 is written |x - 0| or simply |x|, and the distance from x to 5 is |x - 5|. Thus x is farther from 0 than it is from 5 implies |x| > |x - 5| (or, |x - 5| < |x|).

b) x is more than b units from a implies |x - a| > b. x is less than c units from a implies |x - a| < c. Since b < |x - a| and |x - a| < c, we conclude b < |x - a| < c.

Regards,

Rich B.

--- In Math4u@yahoogroups.com, dghbosa alias <redgate3000@...> wrote:
>
> Hi everyone can you help me to solve these questions :
>
> 1.Calculate [ ( -2 ) ^ 0 y ^ -1 ( 2y ) ^ 3 / ( 2y ^ -2 ) ^ -1 y ^ -4 ]
>
> 2. Use the absolute value notation " | | " to discribe :
>
> a. x is farther from 0 than it is from 5 .
>
> b. x is more than b units from a but less than c units from a .
>
>
> ---------------------------------
> Be a better friend, newshound, and know-it-all with Yahoo! Mobile. Try it now.
>

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[Math4u] Re: Cube problem

2007-12-29T13:06:00.001-08:00

OK,
In that case (where for [1,2,3,4] 1 and 2 are opposite, 3 and 4 are
opposite) I believe
[1234] and [1243] and [3412] are duplicates
[1235] and [1253] and [3512] are duplicates
[1236] and [1263] and [3612] are duplicates
For example we can rotate [1,2,3,4] about the 1,2 axis and get
[1,2,4,3]
135 printed out ways divided by 3 duplicates in each group equals 45,
ANSWER.

Another way of looking at the problems:
Question 1, 6 different colors; find the number of ways of picking
unique colors for 1 pair of opposite sides.
Question 2, 6 different colors; find the number of ways of picking
unique colors for 2 pairs of opposite sides.
Question 3, 6 different colors; find the number of ways of picking
unique colors for 3 pairs of opposite sides.

I am feeling more confident in my answers
Question 1 we agree the answer is 6*5/2=15
Question 2
Ways of picking 4 colors, any order = 6*5*4*3/(4*3*2*1)=15
Ways of excluding 2 colors, any order = 6*5/(2*1)=15
Pick any color. It has 3 choices for the opposite color. So the
answer is
15*3=45
Question 3
Method A
Pick any color.
You have 5 choices for the opposite side.
Pick any of the remaining 4 colors.
You have 3 choices for the opposite side
You have 2 choices for the next 2 colors.
5*3*2=30 Answer
Method B
Assign a color to a side.
You have 5*4*3*2*1 or 120 ways of assigning the other colors.
Keeping our chosen color to the front, we can rotate the cube into 4
positions.
120/4=30 Answer
Method C
Don't move the cube
Ways of assigning the colors = 6*5*4*3*2*1=720
There are 6 different faces that can be put in front and for each
face the cube can be rotated into 4 positions so there are 24
orientations of the cube.
720/24=30 answer.
Method D
We take our answer of 45 from question 2. There are 2 ways to add
the last two colors. But we will have three duplicates for each
combination.
45*2/3=30 answer.
Regards,
Brian

--- In Math4u@yahoogroups.com, "w7anf" <cherry@...> wrote:
>
> Sides are numbered 1 bottom, 2 top, 3 left, 4 right, 5 back, 6
front.
> [1,2,3,4] means color 1 is on the bottom.
> Color 2 is opposite on the top.
> Color 3 is on the left.
> Color 4 is opposite on the right.
> [1,4,2,3] is different and can not be rotated to [1,2,3,4].
> Jim FitzSimons
>
> --- In Math4u@yahoogroups.com, "Brian Edward Jensen"
> <brianejensen@> wrote:
> >
> > Jim, you went to a lot of work!
> > My answers are
> > Question 1 = 15, you agree
> > Question 2 = 45, Jim got 135
> > Question 3 = 30
> > I have a feeling I made another mistake. Seems to me that once
you
> > have the ring of 4 colors around the cube, there are only 2 ways
> to
> > assign the remaining 2 colors so answer 3 should be twice answer
> 2.
> > I'll look at it tonight.
> > So Jim's answer is 3 times my answer for question 2. It could be
> > interpretation. I look at Jim's list of 135 possibilities. I
would
> > interpret [1234], [1432], and [3412] as duplicates because they
> each
> > have 1 and 3 opposite. If you divide 135 by 3 you get 45. So Jim
> and
> > I are on the same track.
> > Looking at Jim's solution for question 3, I can't tell which
sides
> > are opposite. If the three pairs of opposites were the same,
there
> > would still be two unique solutions. We could call them right
hand
> or
> > left hand with the thumb on one axis pointing toward increasing
> value
> > and the fingers in the direction of rotation. Don't know how we'd
> > decide the direction.
> >
> > Regards,
> > Brian
> >
> > --- In Math4u@yahoogroups.com, "w7anf" <cherry@> wrote:
> > >
> > > Question:-
> > > Given six different colours and a cube.
> > > 1. In how many ways two opposite faces of the cube can
> > > be coloured with two different colours?
> > > 2. In how many ways four faces of the cube can be
> > > coloured with four different colours of which two
> > > faces are opposite and other two faces are also
> > > opposite?
> > > 3. In how many ways all six faces of the cube can be
> > > coloured with six different colours?
> > >
> > > Colors 1,2,3,4,5,6
> > > Sides 1 bottom, 2 top, 3 left, 4 right, 5 back, 6 front
> > >
> > > 1. C(6,2)=6*5/2=15
> > > [[1,2],[1,3],[1,4],[1,5],[1,6],[2,3],[2,4],[2,5],[2,6],
> > > [3,4],[3,5],[3,6],[4,5],[4,6],[5,6]]
> > >
> > > 2. 15*4*3=180
> > > Flip and rotate.
> > > Without duplicates there are
> > > 135 different ways.
> > > [[1,2,3,4],[1,2,3,5],[1,2,3,6],[1,2,4,3],[1,2,4,5],
> > > [1,2,4,6],[1,2,5,3],[1,2,5,4],[1,2,5,6],[1,2,6,3],
> > > [1,2,6,4],[1,2,6,5],[1,3,2,4],[1,3,2,5],[1,3,2,6],
> > > [1,3,4,2],[1,3,4,5],[1,3,4,6],[1,3,5,2],[1,3,5,4],
> > > [1,3,5,6],[1,3,6,2],[1,3,6,4],[1,3,6,5],[1,4,2,3],
> > > [1,4,2,5],[1,4,2,6],[1,4,3,2],[1,4,3,5],[1,4,3,6],
> > > [1,4,5,2],[1,4,5,3],[1,4,5,6],[1,4,6,2],[1,4,6,3],
> > > [1,4,6,5],[1,5,2,3],[1,5,2,4],[1,5,2,6],[1,5,3,2],
> > > [1,5,3,4],[1,5,3,6],[1,5,4,2],[1,5,4,3],[1,5,4,6],
> > > [1,5,6,2],[1,5,6,3],[1,5,6,4],[1,6,2,3],[1,6,2,4],
> > > [1,6,2,5],[1,6,3,2],[1,6,3,4],[1,6,3,5],[1,6,4,2],
> > > [1,6,4,3],[1,6,4,5],[1,6,5,2],[1,6,5,3],[1,6,5,4],
> > > [2,3,1,4],[2,3,1,5],[2,3,1,6],[2,3,4,5],[2,3,4,6],
> > > [2,3,5,4],[2,3,5,6],[2,3,6,4],[2,3,6,5],[2,4,1,3],
> > > [2,4,1,5],[2,4,1,6],[2,4,3,5],[2,4,3,6],[2,4,5,3],
> > > [2,4,5,6],[2,4,6,3],[2,4,6,5],[2,5,1,3],[2,5,1,4],
> > > [2,5,1,6],[2,5,3,4],[2,5,3,6],[2,5,4,3],[2,5,4,6],
> > > [2,5,6,3],[2,5,6,4],[2,6,1,3],[2,6,1,4],[2,6,1,5],
> > > [2,6,3,4],[2,6,3,5],[2,6,4,3],[2,6,4,5],[2,6,5,3],
> > > [2,6,5,4],[3,4,1,2],[3,4,1,5],[3,4,1,6],[3,4,2,5],
> > > [3,4,2,6],[3,4,5,6],[3,4,6,5],[3,5,1,2],[3,5,1,4],
> > > [3,5,1,6],[3,5,2,4],[3,5,2,6],[3,5,4,6],[3,5,6,4],
> > > [3,6,1,2],[3,6,1,4],[3,6,1,5],[3,6,2,4],[3,6,2,5],
> > > [3,6,4,5],[3,6,5,4],[4,5,1,2],[4,5,1,3],[4,5,1,6],
> > > [4,5,2,3],[4,5,2,6],[4,5,3,6],[4,6,1,2],[4,6,1,3],
> > > [4,6,1,5],[4,6,2,3],[4,6,2,5],[4,6,3,5],[5,6,1,2],
> > > [5,6,1,3],[5,6,1,4],[5,6,2,3],[5,6,2,4],[5,6,3,4]]
> > >
> > > 3. 135*2=270
> > > Flip and rotate.
> > > Without duplicates there are
> > > 210 different ways.
> > > [[1,2,3,4,5,6],[1,2,3,4,6,5],[1,2,3,5,4,6],[1,2,3,5,6,4],
> > > [1,2,3,6,4,5],[1,2,3,6,5,4],[1,2,4,3,5,6],[1,2,4,3,6,5],
> > > [1,2,4,5,3,6],[1,2,4,5,6,3],[1,2,4,6,3,5],[1,2,4,6,5,3],
> > > [1,2,5,3,4,6],[1,2,5,3,6,4],[1,2,5,4,3,6],[1,2,5,4,6,3],
> > > [1,2,5,6,3,4],[1,2,5,6,4,3],[1,2,6,3,4,5],[1,2,6,3,5,4],
> > > [1,2,6,4,3,5],[1,2,6,4,5,3],[1,2,6,5,3,4],[1,2,6,5,4,3],
> > > [1,3,2,4,5,6],[1,3,2,4,6,5],[1,3,2,5,4,6],[1,3,2,5,6,4],
> > > [1,3,2,6,4,5],[1,3,2,6,5,4],[1,3,4,2,5,6],[1,3,4,2,6,5],
> > > [1,3,4,5,2,6],[1,3,4,5,6,2],[1,3,4,6,2,5],[1,3,4,6,5,2],
> > > [1,3,5,2,4,6],[1,3,5,2,6,4],[1,3,5,4,2,6],[1,3,5,4,6,2],
> > > [1,3,5,6,2,4],[1,3,5,6,4,2],[1,3,6,2,4,5],[1,3,6,2,5,4],
> > > [1,3,6,4,2,5],[1,3,6,4,5,2],[1,3,6,5,2,4],[1,3,6,5,4,2],
> > > [1,4,2,3,5,6],[1,4,2,3,6,5],[1,4,2,5,3,6],[1,4,2,5,6,3],
> > > [1,4,2,6,3,5],[1,4,2,6,5,3],[1,4,3,2,5,6],[1,4,3,2,6,5],
> > > [1,4,3,5,2,6],[1,4,3,5,6,2],[1,4,3,6,2,5],[1,4,3,6,5,2],
> > > [1,4,5,2,3,6],[1,4,5,2,6,3],[1,4,5,3,2,6],[1,4,5,3,6,2],
> > > [1,4,5,6,2,3],[1,4,5,6,3,2],[1,4,6,2,3,5],[1,4,6,2,5,3],
> > > [1,4,6,3,2,5],[1,4,6,3,5,2],[1,4,6,5,2,3],[1,4,6,5,3,2],
> > > [1,5,2,3,4,6],[1,5,2,3,6,4],[1,5,2,4,3,6],[1,5,2,4,6,3],
> > > [1,5,2,6,3,4],[1,5,2,6,4,3],[1,5,3,2,4,6],[1,5,3,2,6,4],
> > > [1,5,3,4,2,6],[1,5,3,4,6,2],[1,5,3,6,2,4],[1,5,3,6,4,2],
> > > [1,5,4,2,3,6],[1,5,4,2,6,3],[1,5,4,3,2,6],[1,5,4,3,6,2],
> > > [1,5,4,6,2,3],[1,5,4,6,3,2],[1,5,6,2,3,4],[1,5,6,2,4,3],
> > > [1,5,6,3,2,4],[1,5,6,3,4,2],[1,5,6,4,2,3],[1,5,6,4,3,2],
> > > [1,6,2,3,4,5],[1,6,2,3,5,4],[1,6,2,4,3,5],[1,6,2,4,5,3],
> > > [1,6,2,5,3,4],[1,6,2,5,4,3],[1,6,3,2,4,5],[1,6,3,2,5,4],
> > > [1,6,3,4,2,5],[1,6,3,4,5,2],[1,6,3,5,2,4],[1,6,3,5,4,2],
> > > [1,6,4,2,3,5],[1,6,4,2,5,3],[1,6,4,3,2,5],[1,6,4,3,5,2],
> > > [1,6,4,5,2,3],[1,6,4,5,3,2],[1,6,5,2,3,4],[1,6,5,2,4,3],
> > > [1,6,5,3,2,4],[1,6,5,3,4,2],[1,6,5,4,2,3],[1,6,5,4,3,2],
> > > [2,3,1,4,5,6],[2,3,1,4,6,5],[2,3,1,5,4,6],[2,3,1,5,6,4],
> > > [2,3,1,6,4,5],[2,3,1,6,5,4],[2,3,4,5,1,6],[2,3,4,6,1,5],
> > > [2,3,5,4,1,6],[2,3,5,6,1,4],[2,3,6,4,1,5],[2,3,6,5,1,4],
> > > [2,4,1,3,5,6],[2,4,1,3,6,5],[2,4,1,5,3,6],[2,4,1,5,6,3],
> > > [2,4,1,6,3,5],[2,4,1,6,5,3],[2,4,3,5,1,6],[2,4,3,6,1,5],
> > > [2,4,5,3,1,6],[2,4,5,6,1,3],[2,4,6,3,1,5],[2,4,6,5,1,3],
> > > [2,5,1,3,4,6],[2,5,1,3,6,4],[2,5,1,4,3,6],[2,5,1,4,6,3],
> > > [2,5,1,6,3,4],[2,5,1,6,4,3],[2,5,3,4,1,6],[2,5,3,6,1,4],
> > > [2,5,4,3,1,6],[2,5,4,6,1,3],[2,5,6,3,1,4],[2,5,6,4,1,3],
> > > [2,6,1,3,4,5],[2,6,1,3,5,4],[2,6,1,4,3,5],[2,6,1,4,5,3],
> > > [2,6,1,5,3,4],[2,6,1,5,4,3],[2,6,3,4,1,5],[2,6,3,5,1,4],
> > > [2,6,4,3,1,5],[2,6,4,5,1,3],[2,6,5,3,1,4],[2,6,5,4,1,3],
> > > [3,4,1,2,5,6],[3,4,1,2,6,5],[3,4,1,5,2,6],[3,4,1,6,2,5],
> > > [3,4,2,5,1,6],[3,4,2,6,1,5],[3,4,5,6,1,2],[3,4,6,5,1,2],
> > > [3,5,1,2,4,6],[3,5,1,2,6,4],[3,5,1,4,2,6],[3,5,1,6,2,4],
> > > [3,5,2,4,1,6],[3,5,2,6,1,4],[3,5,4,6,1,2],[3,5,6,4,1,2],
> > > [3,6,1,2,4,5],[3,6,1,2,5,4],[3,6,1,4,2,5],[3,6,1,5,2,4],
> > > [3,6,2,4,1,5],[3,6,2,5,1,4],[3,6,4,5,1,2],[3,6,5,4,1,2],
> > > [4,5,1,2,3,6],[4,5,1,3,2,6],[4,5,1,6,2,3],[4,5,2,3,1,6],
> > > [4,5,2,6,1,3],[4,5,3,6,1,2],[4,6,1,2,3,5],[4,6,1,3,2,5],
> > > [4,6,1,5,2,3],[4,6,2,3,1,5],[4,6,2,5,1,3],[4,6,3,5,1,2],
> > > [5,6,1,2,3,4],[5,6,1,3,2,4],[5,6,1,4,2,3],[5,6,2,3,1,4],
> > > [5,6,2,4,1,3],[5,6,3,4,1,2]]
> > >
> > > Jim FitzSimons
> > >
> > > --- In Math4u@yahoogroups.com, "Brian Edward Jensen"
> > > <brianejensen@> wrote:
> > > >
> > > > Question 1
> > > > I agree with Jim's answer. There are 6 color choices for one
> > > square
> > > > and 5 remaining color choices for the opposite square. But
> the
> > > cube
> > > > can be flipped over causing half the choices to be
> duplicates.
> > So
> > > > the answer is 6*5/2=15
> > > > Question 2
> > > > First of all we chose 4 colors out of 6 different colors.
> > > > I get 6*5*4*3=360.
> > > > Let's chose two colors to exclude:
> > > > I get 6*5=30
> > > > What a discrepancy! We should get the same answer picking 4
> > > colors
> > > > or excluding 2 colors.
> > > > Obviously, if we chose 6 colors out of six different colors,
> > there
> > > is
> > > > only one possibility. So we must chose the lower number.
> There
> > > are
> > > > 30 different ways to choose 4 colors out of 6.
> > > > Let's call these colors 1,2,3,and 4.
> > > > These 4 colors are to be arranged in a band around the cube.
> One
> > > of
> > > > the sides will have color 1. There are three remaining
colors
> to
> > > > choose from to be opposite color 1.
> > > > So I'd say the answer is 30*3=90.
> > > > Question 3
> > > > We have a cube colored with 6 colors.
> > > > There are 5 choices for the color opposite color 1. Let's
> call
> > > the
> > > > remaining colors 3,4,5,and 6. Color 3 will be between the
> first
> > > two
> > > > colors.
> > > > There are 3 choices for the color opposite color 3.
> > > > There are 2 remaining choices for arranging the last 2 colors.
> > > > So I'd say the answer is 5*3*2=30
> > > > Regards,
> > > > Brian
> > > >
> > > >
> > > > --- In Math4u@yahoogroups.com, "sanjivadayal" <sanjivadayal@>
> > > > wrote:
> > > > >
> > > > > Question:-
> > > > > Given six different colours and a cube.
> > > > > 1. In how many ways two opposite faces of the cube can
> > > > > be coloured with two different colours?
> > > > > 2. In how many ways four faces of the cube can be
> > > > > coloured with four different colours of which two
> > > > > faces are opposite and other two faces are also
> > > > > opposite?
> > > > > 3. In how many ways all six faces of the cube can be
> > > > > coloured with six different colours?
> > > > >
> > > >
> > >
> >
>

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[Math4u] Re: Fun Problems

2007-12-29T10:52:00.000-08:00

I have to admit that I had so much fun solving these problems about
coins. From a single problem, which seems to have no systematic way
of solving, it was generalized into a mathematical method.
Thank you Brian,
Thank you other people who participated
Now I wonder if anyone can propose another difficult problem about
coins.

Sinichi7kudo

--- In Math4u@yahoogroups.com, "Brian Edward Jensen"
<brianejensen@...> wrote:
>
> To distribute the coins, the coin labeled 120 in the base 3 number
> system, for example, would be in pan 1 for the first weighing, pan
2
> for the second weighing, and off to the side for the third
weighing,
> hence the "120."
> Yes, there are many solutions because we must make an arbitrary
> choice between label 112 and 221 which are opposites. We must
choose
> one from each pair so that for all the coins, we have the same
number
> of coins on each pan. But with thirteen coins, we wind up with 4
> coins on one pan and 5 coins on the other pan so we must borrow a
> genuine coin.
> I don't know about other methods. You came up with a brilliant
> alternative method for 12 coins.
> Regards,
> Brian
>
> --- In Math4u@yahoogroups.com, "sinichi7kudo" <sinichi7kudo@>
> wrote:
> >
> > To Brian,
> > The coins problem became so beautiful.
> > Surprisingly, I solved it the same way as you did, and I also
> > figured out that 14 coins can be solved with certain
restrictions.
> > But I didn't know how to distribute the coins on the balance and
> how
> > to move them. So according to your solution, are there a lot of
> ways
> > to arrange the coins on the balance?
> > Further more, for the 13 coins, I wonder if we have any solution
to
> > distinguish the fake (don't know if it's lighter or heavier),
> > besides using the method that you did.
> >
> > Regards,
> > Sinichi7kudo
> > --- In Math4u@yahoogroups.com, "Brian Edward Jensen"
> > <brianejensen@> wrote:
> > >
> > > To Sinichi7kudo
> > > I will study your 13 coin solution when you give it. You
caused
> > me
> > > to study my solution further. The basic solution can be done
for
> > 12
> > > coins. We can do it for an additional coin is we leave one
coin
> > off
> > > to the side for all three weighings but we won't know is this
> coin
> > is
> > > heavy or light. We can do an additional coin if we add a
genuine
> > > coin on pan 2 for each of the three weighings so that we test
5
> > coins
> > > on pan 1 and 4 coins on pan 2 each time. This gets us up to
14
> > coins
> > > but we will not know if the unweighed coin is heavy or light.
> > >
> > > Let's say we call the two pans on our balance pan 1 and pan
2.
> > > IDEA: We could have a balance with 3 pans arranged in a
> > triangle.
> > > Group 3 is off to the side and not weighed.
> > >
> > > Let's write all the three digit numbers 0-26 in the base 3
number
> > > system:
> > > 000
> > > 001
> > > 002
> > > 010
> > > 011
> > > 012
> > > 020
> > > etceteras
> > >
> > > Let these 27 numbers be the individual ID numbers of the
coins.
> > > Let the first digit be the first weighing, the second digit be
> the
> > > second weighing and the third digit be the third weighing.
> > > Let the "0" mean the coin is not on the balance.
> > > Let the "1" mean the coin is on pan 1.
> > > Let the "2" mean the coin is on pan 2.
> > > If we know that one of the coins is 25% heavier, we can
identify
> > > which of the 27 coins is the heavy coin. If the balance
> balances,
> > > then we have a zero. If pan 1 goes down, then we have a "1."
If
> > pan
> > > 2 goes down, then we have a "2."
> > >
> > > If we don't know if the bad coin is heavy or light, we will
not
> be
> > > able to distinguish between 001 and 002 or 012 and 021 for
> > examples.
> > > So in this case we must pair up the indistinguishable pairs of
> > > numbers as follows:
> > > 000
> > > 001, 002
> > > 010, 020
> > > 011, 022
> > > 012, 021
> > > 100, 200
> > > 101, 202
> > > 102, 201
> > > 110, 220
> > > 111, 222
> > > 112, 221
> > > 120, 210
> > > 121, 212
> > > 122, 211
> > >
> > > What we need to do next is to choose one of each of the two
> > numbers
> > > so that we have some coins on each pan. There are many
solutions
> > so
> > > just try some choices at random and then make a few switches:
> > >
> > > 000
> > > 001, 002 choose 001
> > > 010, 020 choose 010
> > > 011, 022 choose 011
> > > 012, 021 choose 012
> > > 100, 200 choose 200
> > > 101, 202 choose 101
> > > 102, 201 choose 102
> > > 110, 220 choose 220
> > > 111, 222 choose 111
> > > 112, 221 choose 221
> > > 120, 210 choose 120
> > > 121, 212 choose 212
> > > 122, 211 choose 122
> > >
> > > If we count the 1's in the first column of the 13 choices, we
> find
> > we
> > > have 5 coins in pan 1 and 4 coins in pan 2. The same is true
for
> > the
> > > second and third weighings.
> > >
> > > Options:
> > > If we delete coin 111, then we will have 4 coins in each pan.
> For
> > 12
> > > coins, we can tell if the bad coin is heavy or light.
> > > We can add a thirteenth coin 000 which will be indicated as
the
> > bad
> > > coin if the balance balances all three weighings but we won't
> know
> > if
> > > it is heavy or light.
> > > We can add a genuine coin so that we have 5 coins on each side
> > > instead of 4 on one side and 5 on the other. Now we can tell
> > whether
> > > the bad coin is heavy or light out of 13 coins.
> > > We can add a fourteenth coin 000 which will be indicated as
the
> > bad
> > > coin is the balance balances all three weighings but we won't
> know
> > if
> > > it is heavy or light.
> > >
> > > Regards,
> > > Brian
> > >
> > > --- In Math4u@yahoogroups.com, "sinichi7kudo" <sinichi7kudo@>
> > > wrote:
> > > >
> > > > I just have another question: If you know that the fake coin
is
> > > > heavier than other coins. And you are allowed to do 3
> > weightings.
> > > So
> > > > what is the maximum number of coins?
> > > > And by the way, I found out a way to weight 3 times for 13
> coins
> > to
> > > > find out the fake coin (without knowing the fake is heavier
or
> > > > lighter)
> > > > Thank you.
> > > > --- In Math4u@yahoogroups.com, "Michael S." <M.Suesserott@>
> > wrote:
> > > > >
> > > > > There is a large body of literature on this coin weighing
> > > problem.
> > > > For
> > > > > starters, you might look at
> > > > > http://www.cut-the-knot.org/blue/OddballProblem1.shtml
> > > > > and follow the links given at the end of the article.
> > > > >
> > > > > This problem can also be attacked via an information
> > theoretical
> > > > > approach, using entropy calculations. IMHO, this is one of
> the
> > > > best and
> > > > > most elucidating ways to solve the puzzle. For a brief
> > > > introduction, see
> > > > > http://www.av8n.com/physics/twelve-coins.htm
> > > > >
> > > > > For those who want to delve deeper into this, there is the
> > > > excellent
> > > > > book, "Probability and Information" (Theory and Decision
> > Library)
> > > > by
> > > > > A.M. Yaglom and I.M. Yaglom.
> > > > >
> > > > > Best wishes,
> > > > >
> > > > > Michael
> > > > >
> > > > >
> > > > > Mosaad Alabdullatif wrote:
> > > > > > So great
> > > > > > So magnificent
> > > > > >
> > > > > > I really love it
> > > > > >
> > > > > > The question now: if they were 13 coins, we would be
forced
> > to
> > > > hit the
> > > > > > 6-6 route. Would this way leads to such a briliant
solution?
> > > > > >
> > > > > > Thank you Sinichi7kudo
> > > > > > Thank you Brian
> > > > > >
> > > > > > Mosaad
> > > > > >
> > > > > > */sinichi7kudo <sinichi7kudo@>/* wrote:
> > > > > >
> > > > > > Here is how you do it. It certainly can be solved.
I'm
> > now
> > > > trying to
> > > > > > apply the same logic to more than 12 coins, but no
> > > successes
> > > > so far.
> > > > > >
> > > > > > Divide the coins into three groups of 4 coins each.
> > First,
> > > > weigh two
> > > > > > of the groups. If they balance, then the fake coin
is
> in
> > > the
> > > > third
> > > > > > group, so take three of the coins, and weigh them
> > against
> > > > three of
> > > > > > the coins that we know for sure are real (from the
> first
> > 8
> > > > coins).
> > > > > > If they balance, then the remaining coin is the fake
> > one,
> > > > but we
> > > > > > still don't know if it is lighter or heavier, so
weigh
> > it
> > > > against a
> > > > > > real coin to determine that. But if it didn't
balance,
> > then
> > > > we know
> > > > > > that the fake coin is one of the three, and we also
> know
> > if
> > > > it is
> > > > > > lighter or heavier, because we know which side had
the
> > real
> > > > coins.
> > > > > > So weigh two of the coins against each other, if
they
> > > > balance then
> > > > > > the third coin is fake, if they don't balance we
know
> > which
> > > > coin is
> > > > > > fake, because we already know if it is lighter or
> > heavier.
> > > > > > Now let's go back to the first weighing; if the two
> > groups
> > > > didn't
> > > > > > balance, we know that the fake coin is in one of the
> > > groups,
> > > > but we
> > > > > > don't know which group. Let's mark the group that
was
> > > > heavier as
> > > > > > group A, and the group that was lighter group B. For
> the
> > > > second
> > > > > > weighing, weigh two coins of group A and one coin of
> > group
> > > B,
> > > > > > against the other two coins of group A and one coin
of
> > > group
> > > > B. If
> > > > > > they balance, the fake coin must be one of the
> remaining
> > > > two, and a
> > > > > > weigh between them will tell us which one (they are
> both
> > > > from group
> > > > > > B so we know that it's lighter). If the second weigh
> > didn't
> > > > balance,
> > > > > > look at the side that is heavier: Either the fake
coin
> > is
> > > > one of the
> > > > > > two group A coins, or it can the group B coin on the
> > other
> > > > side of
> > > > > > the scale. Take the two group A coins, and weigh
them
> > > > against each
> > > > > > other. If they balance, then it is the group B coin
> that
> > is
> > > > fake,
> > > > > > and if they don't balance, then the heavier one is
the
> > fake
> > > > coin.
> > > > > > And that's that. We have covered all possibilities.
> > > > > >
> > > > > > --- In Math4u@yahoogroups.com <mailto:Math4u%
> > > > 40yahoogroups.com>,
> > > > > > BRIAN JENSEN <brianejensen@> wrote:
> > > > > > >
> > > > > > > My sixth grade teacher gave this problem to us
around
> > > > 1962. It
> > > > > > took me 38 years to figure out. He didn't remember
me
> or
> > > the
> > > > problem.
> > > > > > > I have found many solutions to this problem. There
is
> > no
> > > > regular
> > > > > > pattern that I could find. If you can put the coins
on
> > the
> > > > left pan,
> > > > > > right pan, or to the side, then how many different
ways
> > are
> > > > there to
> > > > > > position the coins in three weighings? It certainly
can
> > be
> > > > solved.
> > > > > > > Regards,
> > > > > > > Brian
> > > > > > >
> > > > > > >
> > > > > > > Mosaad Alabdullatif <sam362yah@> wrote:
> > > > > > > I do not think 3 wieghings will be enough if we do
not
> > > > > > know if that defected coin is light or heavy!!
> > > > > > >
> > > > > > > Mosaad
> > > > > > >
> > > > > > > Brian Edward Jensen <brianejensen@> wrote:
> > > > > > > I made a mistake! (Stephen Tavener pointed this
out
> to
> > > me.)
> > > > > > On the
> > > > > > > second problem, I meant to say "sphere,"
not "cube."
> I
> > > > started
> > > > > > > writing this problem thinking that the area of a
cube
> > > > could be
> > > > > > > obtained from derivative of the volume with
respect
> to
> > > > half a
> > > > > > side.
> > > > > > > But this would be too easy because someone could
> > easily
> > > > calculate
> > > > > > the
> > > > > > > area of a side and multiply by 6 sides for the
total
> > > area.
> > > > So then
> > > > > > I
> > > > > > > decided to use a sphere.
> > > > > > >
> > > > > > > Here is a resubmission of the two problems and I
am
> > > > deleting the
> > > > > > > original:
> > > > > > >
> > > > > > > Problem 1:
> > > > > > > Here's an old problem, one of my favorites that we
> > > haven't
> > > > had in
> > > > > > a
> > > > > > > couple years:
> > > > > > > A man has 12 coins and one is bad. So eleven coins
> > have
> > > > the same
> > > > > > > weight and one has a different weight. The man
puts
> > them
> > > > on a
> > > > > > > balance, some coins on each pan. He can make 3
> > weighings.
> > > > This is
> > > > > > an
> > > > > > > old-fashioned balance with 2 pans on opposite ends
of
> > an
> > > > arm. Of
> > > > > > > course he must have the same number of coins on
each
> > side
> > > > but he
> > > > > > > doesn't need to weigh all the coins each time.
> > > > > > > How can he figure out which coin is bad with 3
> > weighings
> > > > if he
> > > > > > knows
> > > > > > > the bad coin is heavy? (Actually, he can pick the
bad
> > one
> > > > out of
> > > > > > 27
> > > > > > > coins if he knows one is 25% heavy)
> > > > > > > Alternately and far more difficult,
> > > > > > > How can he figure out which coin is bad with 3
> > weighings
> > > > if it is
> > > > > > not
> > > > > > > known if the coin is heavy or light? (Actually
> Stephen
> > > > Tavener is
> > > > > > > correct in a private communication saying 13 coins
> > would
> > > be
> > > > > > > possible. If a 13th coin is left on the side
without
> > > > weighing, and
> > > > > > a
> > > > > > > bad coin is not identified among the twelve, we
would
> > > know
> > > > it is
> > > > > > coin
> > > > > > > 13. We would not know if this coin were heavy or
> > light,
> > > > however.)
> > > > > > >
> > > > > > > Problem 2:
> > > > > > > Another problem:
> > > > > > > If the volume of a sphere is C*r^3
> > > > > > > Where
> > > > > > > C is a constant
> > > > > > > r is the radius
> > > > > > > Using simple differentiation instead of looking it
up
> > in
> > > a
> > > > book,
> > > > > > what
> > > > > > > is the
> > > > > > > surface area of the sphere? Show your logic.
> > > > > > > Regards,
> > > > > > > Brian
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > ---------------------------------
> > > > > > > Looking for last minute shopping deals? Find them
> fast
> > > with
> > > > > > Yahoo! Search.
> > > > > > >
> > > > > >
> > > > > >
> > > > > > ---------------------------------------------------------
---
> -
> > ---
> > > -
> > > > -------
> > > > > > Be a better friend, newshound, and know-it-all with
Yahoo!
> > > > Mobile. Try
> > > > > > it now.
> > > > > >
> > > >
> > >
> >
>
<http://us.rd.yahoo.com/evt=51733/*http://mobile.yahoo.com/;_ylt=Ahu0
> > > > 6i62sR8HDtDypao8Wcj9tAcJ%20>
> > > > > >
> > > > >
> > > >
> > >
> >
>

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[Math4u] Re: Cube problem

2007-12-29T07:52:00.001-08:00

Sides are numbered 1 bottom, 2 top, 3 left, 4 right, 5 back, 6 front.
[1,2,3,4] means color 1 is on the bottom.
Color 2 is opposite on the top.
Color 3 is on the left.
Color 4 is opposite on the right.
[1,4,2,3] is different and can not be rotated to [1,2,3,4].
Jim FitzSimons

--- In Math4u@yahoogroups.com, "Brian Edward Jensen"
<brianejensen@...> wrote:
>
> Jim, you went to a lot of work!
> My answers are
> Question 1 = 15, you agree
> Question 2 = 45, Jim got 135
> Question 3 = 30
> I have a feeling I made another mistake. Seems to me that once you
> have the ring of 4 colors around the cube, there are only 2 ways
to
> assign the remaining 2 colors so answer 3 should be twice answer
2.
> I'll look at it tonight.
> So Jim's answer is 3 times my answer for question 2. It could be
> interpretation. I look at Jim's list of 135 possibilities. I would
> interpret [1234], [1432], and [3412] as duplicates because they
each
> have 1 and 3 opposite. If you divide 135 by 3 you get 45. So Jim
and
> I are on the same track.
> Looking at Jim's solution for question 3, I can't tell which sides
> are opposite. If the three pairs of opposites were the same, there
> would still be two unique solutions. We could call them right hand
or
> left hand with the thumb on one axis pointing toward increasing
value
> and the fingers in the direction of rotation. Don't know how we'd
> decide the direction.
>
> Regards,
> Brian
>
> --- In Math4u@yahoogroups.com, "w7anf" <cherry@> wrote:
> >
> > Question:-
> > Given six different colours and a cube.
> > 1. In how many ways two opposite faces of the cube can
> > be coloured with two different colours?
> > 2. In how many ways four faces of the cube can be
> > coloured with four different colours of which two
> > faces are opposite and other two faces are also
> > opposite?
> > 3. In how many ways all six faces of the cube can be
> > coloured with six different colours?
> >
> > Colors 1,2,3,4,5,6
> > Sides 1 bottom, 2 top, 3 left, 4 right, 5 back, 6 front
> >
> > 1. C(6,2)=6*5/2=15
> > [[1,2],[1,3],[1,4],[1,5],[1,6],[2,3],[2,4],[2,5],[2,6],
> > [3,4],[3,5],[3,6],[4,5],[4,6],[5,6]]
> >
> > 2. 15*4*3=180
> > Flip and rotate.
> > Without duplicates there are
> > 135 different ways.
> > [[1,2,3,4],[1,2,3,5],[1,2,3,6],[1,2,4,3],[1,2,4,5],
> > [1,2,4,6],[1,2,5,3],[1,2,5,4],[1,2,5,6],[1,2,6,3],
> > [1,2,6,4],[1,2,6,5],[1,3,2,4],[1,3,2,5],[1,3,2,6],
> > [1,3,4,2],[1,3,4,5],[1,3,4,6],[1,3,5,2],[1,3,5,4],
> > [1,3,5,6],[1,3,6,2],[1,3,6,4],[1,3,6,5],[1,4,2,3],
> > [1,4,2,5],[1,4,2,6],[1,4,3,2],[1,4,3,5],[1,4,3,6],
> > [1,4,5,2],[1,4,5,3],[1,4,5,6],[1,4,6,2],[1,4,6,3],
> > [1,4,6,5],[1,5,2,3],[1,5,2,4],[1,5,2,6],[1,5,3,2],
> > [1,5,3,4],[1,5,3,6],[1,5,4,2],[1,5,4,3],[1,5,4,6],
> > [1,5,6,2],[1,5,6,3],[1,5,6,4],[1,6,2,3],[1,6,2,4],
> > [1,6,2,5],[1,6,3,2],[1,6,3,4],[1,6,3,5],[1,6,4,2],
> > [1,6,4,3],[1,6,4,5],[1,6,5,2],[1,6,5,3],[1,6,5,4],
> > [2,3,1,4],[2,3,1,5],[2,3,1,6],[2,3,4,5],[2,3,4,6],
> > [2,3,5,4],[2,3,5,6],[2,3,6,4],[2,3,6,5],[2,4,1,3],
> > [2,4,1,5],[2,4,1,6],[2,4,3,5],[2,4,3,6],[2,4,5,3],
> > [2,4,5,6],[2,4,6,3],[2,4,6,5],[2,5,1,3],[2,5,1,4],
> > [2,5,1,6],[2,5,3,4],[2,5,3,6],[2,5,4,3],[2,5,4,6],
> > [2,5,6,3],[2,5,6,4],[2,6,1,3],[2,6,1,4],[2,6,1,5],
> > [2,6,3,4],[2,6,3,5],[2,6,4,3],[2,6,4,5],[2,6,5,3],
> > [2,6,5,4],[3,4,1,2],[3,4,1,5],[3,4,1,6],[3,4,2,5],
> > [3,4,2,6],[3,4,5,6],[3,4,6,5],[3,5,1,2],[3,5,1,4],
> > [3,5,1,6],[3,5,2,4],[3,5,2,6],[3,5,4,6],[3,5,6,4],
> > [3,6,1,2],[3,6,1,4],[3,6,1,5],[3,6,2,4],[3,6,2,5],
> > [3,6,4,5],[3,6,5,4],[4,5,1,2],[4,5,1,3],[4,5,1,6],
> > [4,5,2,3],[4,5,2,6],[4,5,3,6],[4,6,1,2],[4,6,1,3],
> > [4,6,1,5],[4,6,2,3],[4,6,2,5],[4,6,3,5],[5,6,1,2],
> > [5,6,1,3],[5,6,1,4],[5,6,2,3],[5,6,2,4],[5,6,3,4]]
> >
> > 3. 135*2=270
> > Flip and rotate.
> > Without duplicates there are
> > 210 different ways.
> > [[1,2,3,4,5,6],[1,2,3,4,6,5],[1,2,3,5,4,6],[1,2,3,5,6,4],
> > [1,2,3,6,4,5],[1,2,3,6,5,4],[1,2,4,3,5,6],[1,2,4,3,6,5],
> > [1,2,4,5,3,6],[1,2,4,5,6,3],[1,2,4,6,3,5],[1,2,4,6,5,3],
> > [1,2,5,3,4,6],[1,2,5,3,6,4],[1,2,5,4,3,6],[1,2,5,4,6,3],
> > [1,2,5,6,3,4],[1,2,5,6,4,3],[1,2,6,3,4,5],[1,2,6,3,5,4],
> > [1,2,6,4,3,5],[1,2,6,4,5,3],[1,2,6,5,3,4],[1,2,6,5,4,3],
> > [1,3,2,4,5,6],[1,3,2,4,6,5],[1,3,2,5,4,6],[1,3,2,5,6,4],
> > [1,3,2,6,4,5],[1,3,2,6,5,4],[1,3,4,2,5,6],[1,3,4,2,6,5],
> > [1,3,4,5,2,6],[1,3,4,5,6,2],[1,3,4,6,2,5],[1,3,4,6,5,2],
> > [1,3,5,2,4,6],[1,3,5,2,6,4],[1,3,5,4,2,6],[1,3,5,4,6,2],
> > [1,3,5,6,2,4],[1,3,5,6,4,2],[1,3,6,2,4,5],[1,3,6,2,5,4],
> > [1,3,6,4,2,5],[1,3,6,4,5,2],[1,3,6,5,2,4],[1,3,6,5,4,2],
> > [1,4,2,3,5,6],[1,4,2,3,6,5],[1,4,2,5,3,6],[1,4,2,5,6,3],
> > [1,4,2,6,3,5],[1,4,2,6,5,3],[1,4,3,2,5,6],[1,4,3,2,6,5],
> > [1,4,3,5,2,6],[1,4,3,5,6,2],[1,4,3,6,2,5],[1,4,3,6,5,2],
> > [1,4,5,2,3,6],[1,4,5,2,6,3],[1,4,5,3,2,6],[1,4,5,3,6,2],
> > [1,4,5,6,2,3],[1,4,5,6,3,2],[1,4,6,2,3,5],[1,4,6,2,5,3],
> > [1,4,6,3,2,5],[1,4,6,3,5,2],[1,4,6,5,2,3],[1,4,6,5,3,2],
> > [1,5,2,3,4,6],[1,5,2,3,6,4],[1,5,2,4,3,6],[1,5,2,4,6,3],
> > [1,5,2,6,3,4],[1,5,2,6,4,3],[1,5,3,2,4,6],[1,5,3,2,6,4],
> > [1,5,3,4,2,6],[1,5,3,4,6,2],[1,5,3,6,2,4],[1,5,3,6,4,2],
> > [1,5,4,2,3,6],[1,5,4,2,6,3],[1,5,4,3,2,6],[1,5,4,3,6,2],
> > [1,5,4,6,2,3],[1,5,4,6,3,2],[1,5,6,2,3,4],[1,5,6,2,4,3],
> > [1,5,6,3,2,4],[1,5,6,3,4,2],[1,5,6,4,2,3],[1,5,6,4,3,2],
> > [1,6,2,3,4,5],[1,6,2,3,5,4],[1,6,2,4,3,5],[1,6,2,4,5,3],
> > [1,6,2,5,3,4],[1,6,2,5,4,3],[1,6,3,2,4,5],[1,6,3,2,5,4],
> > [1,6,3,4,2,5],[1,6,3,4,5,2],[1,6,3,5,2,4],[1,6,3,5,4,2],
> > [1,6,4,2,3,5],[1,6,4,2,5,3],[1,6,4,3,2,5],[1,6,4,3,5,2],
> > [1,6,4,5,2,3],[1,6,4,5,3,2],[1,6,5,2,3,4],[1,6,5,2,4,3],
> > [1,6,5,3,2,4],[1,6,5,3,4,2],[1,6,5,4,2,3],[1,6,5,4,3,2],
> > [2,3,1,4,5,6],[2,3,1,4,6,5],[2,3,1,5,4,6],[2,3,1,5,6,4],
> > [2,3,1,6,4,5],[2,3,1,6,5,4],[2,3,4,5,1,6],[2,3,4,6,1,5],
> > [2,3,5,4,1,6],[2,3,5,6,1,4],[2,3,6,4,1,5],[2,3,6,5,1,4],
> > [2,4,1,3,5,6],[2,4,1,3,6,5],[2,4,1,5,3,6],[2,4,1,5,6,3],
> > [2,4,1,6,3,5],[2,4,1,6,5,3],[2,4,3,5,1,6],[2,4,3,6,1,5],
> > [2,4,5,3,1,6],[2,4,5,6,1,3],[2,4,6,3,1,5],[2,4,6,5,1,3],
> > [2,5,1,3,4,6],[2,5,1,3,6,4],[2,5,1,4,3,6],[2,5,1,4,6,3],
> > [2,5,1,6,3,4],[2,5,1,6,4,3],[2,5,3,4,1,6],[2,5,3,6,1,4],
> > [2,5,4,3,1,6],[2,5,4,6,1,3],[2,5,6,3,1,4],[2,5,6,4,1,3],
> > [2,6,1,3,4,5],[2,6,1,3,5,4],[2,6,1,4,3,5],[2,6,1,4,5,3],
> > [2,6,1,5,3,4],[2,6,1,5,4,3],[2,6,3,4,1,5],[2,6,3,5,1,4],
> > [2,6,4,3,1,5],[2,6,4,5,1,3],[2,6,5,3,1,4],[2,6,5,4,1,3],
> > [3,4,1,2,5,6],[3,4,1,2,6,5],[3,4,1,5,2,6],[3,4,1,6,2,5],
> > [3,4,2,5,1,6],[3,4,2,6,1,5],[3,4,5,6,1,2],[3,4,6,5,1,2],
> > [3,5,1,2,4,6],[3,5,1,2,6,4],[3,5,1,4,2,6],[3,5,1,6,2,4],
> > [3,5,2,4,1,6],[3,5,2,6,1,4],[3,5,4,6,1,2],[3,5,6,4,1,2],
> > [3,6,1,2,4,5],[3,6,1,2,5,4],[3,6,1,4,2,5],[3,6,1,5,2,4],
> > [3,6,2,4,1,5],[3,6,2,5,1,4],[3,6,4,5,1,2],[3,6,5,4,1,2],
> > [4,5,1,2,3,6],[4,5,1,3,2,6],[4,5,1,6,2,3],[4,5,2,3,1,6],
> > [4,5,2,6,1,3],[4,5,3,6,1,2],[4,6,1,2,3,5],[4,6,1,3,2,5],
> > [4,6,1,5,2,3],[4,6,2,3,1,5],[4,6,2,5,1,3],[4,6,3,5,1,2],
> > [5,6,1,2,3,4],[5,6,1,3,2,4],[5,6,1,4,2,3],[5,6,2,3,1,4],
> > [5,6,2,4,1,3],[5,6,3,4,1,2]]
> >
> > Jim FitzSimons
> >
> > --- In Math4u@yahoogroups.com, "Brian Edward Jensen"
> > <brianejensen@> wrote:
> > >
> > > Question 1
> > > I agree with Jim's answer. There are 6 color choices for one
> > square
> > > and 5 remaining color choices for the opposite square. But
the
> > cube
> > > can be flipped over causing half the choices to be
duplicates.
> So
> > > the answer is 6*5/2=15
> > > Question 2
> > > First of all we chose 4 colors out of 6 different colors.
> > > I get 6*5*4*3=360.
> > > Let's chose two colors to exclude:
> > > I get 6*5=30
> > > What a discrepancy! We should get the same answer picking 4
> > colors
> > > or excluding 2 colors.
> > > Obviously, if we chose 6 colors out of six different colors,
> there
> > is
> > > only one possibility. So we must chose the lower number.
There
> > are
> > > 30 different ways to choose 4 colors out of 6.
> > > Let's call these colors 1,2,3,and 4.
> > > These 4 colors are to be arranged in a band around the cube.
One
> > of
> > > the sides will have color 1. There are three remaining colors
to
> > > choose from to be opposite color 1.
> > > So I'd say the answer is 30*3=90.
> > > Question 3
> > > We have a cube colored with 6 colors.
> > > There are 5 choices for the color opposite color 1. Let's
call
> > the
> > > remaining colors 3,4,5,and 6. Color 3 will be between the
first
> > two
> > > colors.
> > > There are 3 choices for the color opposite color 3.
> > > There are 2 remaining choices for arranging the last 2 colors.
> > > So I'd say the answer is 5*3*2=30
> > > Regards,
> > > Brian
> > >
> > >
> > > --- In Math4u@yahoogroups.com, "sanjivadayal" <sanjivadayal@>
> > > wrote:
> > > >
> > > > Question:-
> > > > Given six different colours and a cube.
> > > > 1. In how many ways two opposite faces of the cube can
> > > > be coloured with two different colours?
> > > > 2. In how many ways four faces of the cube can be
> > > > coloured with four different colours of which two
> > > > faces are opposite and other two faces are also
> > > > opposite?
> > > > 3. In how many ways all six faces of the cube can be
> > > > coloured with six different colours?
> > > >
> > >
> >
>

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[forwards4all] New Year Greetings......

2007-12-29T06:02:00.000-08:00

http://groups.yahoo.com/group/jitmohan/

Welcome To Jitmohan Yahoo Group" in this Group Members can post their messages . And members are requested to send some nice emails, pictures,movie and other type of funny or uncommon things. : Please Join this group : Regards; Jitmohan Group Team / (Owner) Mr,Jitmohan Jha (Jitu) mrjitmohan@yahoo.com , jitu030306@yahoo.co.in , (Moderator) Mr,Ranu Varma ranu030306@yahoo.co.in , 2nd (Mod) Mamta Jha, mamtajha1@yahoo.co.in / Thankyou Have a Nice fun !!

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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ RULES OF THE GROUP: 1. COMMUNAL, ADULT STUFF AND RELATED CONTENT WILL NOT BE ENTERTAINED. Member found posting 'communal' and 'adult' stuff will be simply thrown out of the group. So be careful! 2. DO NOT SPAM OTHER MEMBERS OF THE GROUP BY POSTING UNWANTED EMAILS. GROUP EMAIL ADDRESSES: Groups Home: http://in.groups.yahoo.com/group/forwards4all/ Post message: forwards4all@yahoogroups.co.in Subscribe: forwards4all-subscribe@yahoogroups.co.in Unsubscribe: forwards4all-unsubscribe@yahoogroups.co.in List owner: forwards4all-owner@yahoogroups.co.in ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ *** YAHOO! GROUPS MODERATORS UNION 2007 *** Are you a Moderator or Owner of any Yahoo! Group? Register here for Moderators Union 2007: http://in.promos.yahoo.com/moderators-union/index.html See Yahoo! Groups Moderators blog here: http://blog.360.yahoo.com/moderators_league Check photos of the last event here: http://www.flickr.com/photos/moderators-union/show ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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[forwards4all] School Joke

2007-12-29T05:43:00.000-08:00

A first-grade teacher, Ms Neelam (Age 28) was having trouble with one of her students the teacher asked,"Boy. what is your problem?"

Boy answered, "I'm too smart for the first-grade. My sister is in the third-grade and I'm smarter than she is! I think I should be in the third-grade too!"

Ms Neelam had enough. She took Boy. to the principal's office. While Boy. waited in the outer office, the teacher explained to the principal what the situation was. The principal told Ms Neelam he would give the boy a test and if he failed to answer any of his questions he was to go back to the first-grade and behave.She agreed.

Boy. was brought in and the conditions were explained to him and he agreed to take the test.
Principal: "What is 3 x 3?"

Boy.: "9".
Principal: "What is 6 x 6?"

Boy.: "36".
And so it went with every question the principal thought a third-grade should know. The principal looks at Ms Neelam and tells her, "I think Boy can go to the third-grade."

Ms Neelam says to the principal, "I have some of my own questions.

Can I ask him ?" The principal and Boy both agreed.

Ms Neelam asks, "What does a cow have four of that I have only two of?
Boy., after a moment "Legs."

Ms Nee lam: "What is in your pants that you have but I do not have?"
Boy.: "Pockets."

Ms Neelam: What starts with a C and ends with a T, is hairy, oval, delicious and contains thin whitish liquid?
Boy.: Coconut

Ms Neelam: What goes in hard and pink then comes out soft And sticky? The principal's eyes open really wide and before he could stop the answer, Boy. was taking charge.
Boy.: Bubblegum

Ms Neelam: What does a man do standing up, a woman does sitting down and a dog does on three legs? The principal's eyes open really wide and before he could stop the answer...
Boy.: Shake hands

Ms Neelam: Now I will ask some "Who am I" sort of questions, okay?

Boy.: Yep.

Ms Neelam: You stick your poles inside me. You tie me down to get me up. I get wet before you do.
Boy.: Tent

Ms Neelam: A finger goes in me. You fiddle with me when you're bored. The best man always has me first.The Principal was looking restless, a bit tense and took one large Patiala Vodka peg.
Boy.: Wedding Ring

Ms Neelam: I come in many sizes. When I'm not well, I drip. When you blow me, you feel good.
Boy.: Nose

Ms Neelam: I have a stiff shaft. My tip penetrates. I come with a quiver.
Boy.: Arrow

Ms Neelam: What word sta rts with a 'F' and ends in 'K' that means lot of heat and excitement?
Boy.: Firetruck

Ms Neelam: What word starts with a 'F' and ends in 'K' & if u dont get it u have to use ur hand.
Boy.: Fork

Ms Neelam: What is it that all men have one of it's longer on some men than on others, the pope doesn't use his and a man gives it to his wife after they're married?
Boy.: SURNAME

Ms Neelam: What part of the man has no bone but has muscles, has lots of veins, like pumping, & is responsible for making love ?
Boy.: HEART.

The principal breathed a sigh of relief and said to the teacher...

"Send this Boy to IIM Ahmedabad, I got the last ten questions wrong myself!! "

__,_._,___

[forwards4all] HR Manager's Love Letter

2007-12-29T05:41:00.000-08:00

Dearest Ms Juliet,

I am very happy to inform you that I have fallen in Love with you since the 5th of November (Sunday). With reference to the meeting held between us on the 27th of October at 1500 hrs, I would like to present myself as a prospective lover.

Our love affair would be on probation for a period of three months and depending on compatibility, would be made permanent. Of course, upon completion of probation, there will be continuous on the job training and performance appraisal schemes leading up to promotion from lover to spouse.

The expenses incurred for coffee and entertainment would initially be shared equally between us. Later, based on your performance, I might take up a larger share of the expenses. However I am broadminded enough to be taken care of, on your expense account.

I request you to kindly respond within 30 days of receiving this letter, failing which, this offer would be canceled without further notice and I shall be considering someone else. I would be happy, if you could forward this letter to your sister, if you do not wish to take up this offer.

Thanking you in anticipation,

Yours sincerely,
( HR Executive )

__,_._,___

[forwards4all] Happy New Year 2008.......special card 4 u

2007-12-29T05:40:00.000-08:00

I wish you Happiness and Joy...
And Blessings for the New Year...
I wish you the best of everything...
That you so well deserve...in coming years.

A very special 2008 Greeting Card 4 You

HAPPY NEW YEAR FRIEND!

Yours Sweety!!

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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ RULES OF THE GROUP: 1. COMMUNAL, ADULT STUFF AND RELATED CONTENT WILL NOT BE ENTERTAINED. Member found posting 'communal' and 'adult' stuff will be simply thrown out of the group. So be careful! 2. DO NOT SPAM OTHER MEMBERS OF THE GROUP BY POSTING UNWANTED EMAILS. GROUP EMAIL ADDRESSES: Groups Home: http://in.groups.yahoo.com/group/forwards4all/ Post message: forwards4all@yahoogroups.co.in Subscribe: forwards4all-subscribe@yahoogroups.co.in Unsubscribe: forwards4all-unsubscribe@yahoogroups.co.in List owner: forwards4all-owner@yahoogroups.co.in ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ *** YAHOO! GROUPS MODERATORS UNION 2007 *** Are you a Moderator or Owner of any Yahoo! Group? Register here for Moderators Union 2007: http://in.promos.yahoo.com/moderators-union/index.html See Yahoo! Groups Moderators blog here: http://blog.360.yahoo.com/moderators_league Check photos of the last event here: http://www.flickr.com/photos/moderators-union/show ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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[forwards4all] Chinese Call-Center

2007-12-29T05:37:00.000-08:00

" Good Wan! (Good One!)"

Caller: Hello, can I speak to Annie Wan ?

Operator: Yes, you can speak to me.

Caller : No, I want to speak to Annie Wan!

Operator: Yes I understand you want to speak to anyone.You can speak to me. Who is this?

Caller : I'm Sam Wan. And I need to talk to Annie Wan! It's urgent.

Operator: I know you are someone and you want to talk to anyone! But what's this urgent matter about?

Caller: Well... just tell my sister Annie Wan that our brother Noe Wan was involved in an accident. Noe Wan got injured and now Noe Wan is being sent to the hospital. Right now, Avery Wan is on his way to the hospital.

Operator : Look, if no one was injured and no one was sent to the hospital, then the accident isn't an urgent matter! You may find this hilarious but I don't have time for this!

Caller : You are so rude! Who are you?

Operator: I'm Saw Ree.

Caller: Yes! You should be sorry. Now give me your name!!

Operator: That's what I said. I'm Saw Ree ..

Caller: Oh ......God!!!

Hehehehehe.. :)) :)) :)) :))

__,_._,___

Babbler's Den

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:::|Sweet Jokes |::: Happy New Year 2008..........

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Re: [Math4u] Re: i am in desperate need of help with math

[Math4u] Re: A combinations/probability problem

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Re: [Math4u] A combinations/probability problem

[Math4u] Re: i am in desperate need of help with math

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[Math4u] Re: Questions ???

[Math4u] i am in desperate need of help with math

:::|Sweet Jokes |::: WISH U HAPPY NEW YEAR 2008....

[Inside AdSense] A look back at 2007

:::|Sweet Jokes |::: Happy New Year 2008 !!!!!!!!!!!!!!!!!!!!!!!!!!

[forwards4all] -:|:- Beauty of Snow -:|:-

[Math4u] Leaving Vegas

[forwards4all] Creative Ads !!! Funny Enough :)

[forwards4all] New Year Wishes 4 U !!!

[forwards4all] PENALTY CORNER FOR EVERY MISTAKE !!!

[forwards4all] Happy New Year - 2008

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[forwards4all] !*!~* Cute Cute New Ye@r Gifs *~!*!

[forwards4all] +++ Happy New Year 2008+++

[Math4u] What it could be named?

[Math4u] A combinations/probability problem

[forwards4all] In The New Year

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[forwards4all] True Friendship

[forwards4all] The Best Pictures of 2007

[Math4u] Another problem about balance

[forwards4all] New Speed Limit Device

Re: [Math4u] Re: Cube problem

[forwards4all] Fwd: MICROWAVE

[forwards4all] Stall conversation...

[forwards4all] Australia won 1st Test against India :(

[forwards4all] COMMUNAL JOKES NOT ALLOWED

[forwards4all] Life in pictures -Daughter of the east--BENAZIR BHUTTO

[forwards4all] LEARN THE ART OF SPREADING JEALOUSY !!!

[forwards4all] Amazing Car Stunts

[forwards4all] DHONI YUVRAJ : THE INSIDE STORY !!!!

[Math4u] Re: Questions ???

[Math4u] Re: Cube problem

[Math4u] Re: Fun Problems

[Math4u] Re: Cube problem

[forwards4all] New Year Greetings......

[forwards4all] School Joke

[forwards4all] HR Manager's Love Letter

[forwards4all] Happy New Year 2008.......special card 4 u

[forwards4all] Chinese Call-Center

:::|Sweet Jokes |::: Free Car Wash ( funny )

:::|Sweet Jokes |::: HAPPY NEW YEAR

:::|Sweet Jokes |::: Happy New Year 2008

:::|Sweet Jokes |::: Happy New Year 2008

:::|Sweet Jokes |::: Happy New Year 2008..........

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:::|Sweet Jokes |::: WISH U HAPPY NEW YEAR 2008....

:::|Sweet Jokes |::: Happy New Year 2008 !!!!!!!!!!!!!!!!!!!!!!!!!!

[forwards4all] !!~ New Year Greetings For My Angel friends~!!

[forwards4all] !!~ Cute Cute New Ye@r Gifs ~!!

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