Showing posts with label Math. Show all posts
Showing posts with label Math. Show all posts

Monday, December 31, 2007

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

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

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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Re: [Math4u] A combinations/probability problem

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

--- 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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Re: [Math4u] Another problem about balance

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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[Math4u] Re: Questions ???

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

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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[Math4u] Leaving Vegas

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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Sunday, December 30, 2007

[Math4u] What it could be named?

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


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[Math4u] A combinations/probability problem

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] Another problem about balance

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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Re: [Math4u] Re: Cube problem

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)
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Saturday, December 29, 2007

[Math4u] Re: Questions ???

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

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

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

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] Questions ???

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 .


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Friday, December 28, 2007

[Math4u] a request from an engineer

please guys i'm a student in engineering faculty in Egypt. i'd like to
tell u a suggestion. why don't u care more of calculus;
differentiation and integration? by sending laws and solved or not
solved problems.
i'm waiting for ur kind reply
Urs, eng_ahmf


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[Math4u] Re: Cube problem

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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