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?
>
Brian's answer, Question 2 corrected, 1&3 unchanged below:
I regret that my previous answer for question 2 is high by a factor
of 2. Let's look at the number of ways to reduce 6 different colors
to 4 different colors in any order. This was where I went wrong.
There are two ways of solving this and we should get the same answer
either way. We can pick 4 colors from the 6 colors or we can delete 2
colors from the 6 colors.
Let's say we pick 4 colors.
The number of ways we can pick them in order = 6*5*4*3
The number of ways they can be arranged = 4*3*2*1 which is the number
of duplicates for each selection of 4 colors.
So the number of combinations of four colors can be picked in any
order = 6*5*4*3/(4*3*2*1)
=15
Let's say we delete 2 colors from the 6.
The ways to do it in a specific order = 6*5
The ways to arrange the two colors deleted = 2*1
The number of combinations of 2 colors deleted in any order = 6*5/
(2*1)=15
Notice that this 15 shows up 2 places on Pascal's triangle.
We have 15 different combinations of 4 colors. We arrange them into a
ring. There are 3 different colors that can be opposite a chosen
color. So there are only 3 ways to arrange the 4 colors into a ring.
So the answer is 15*3=45 answer
Question 2 alternate method, same answer
We have 6 different colors. How many different ways can we arrange
them in the ring of 4 squares around the perimeter of a cube omitting
two opposite faces? If we fix the cube, choose a starting square, and
go clockwise, there are
6*5*4*3 different choice possibilities.
The cube can be rotated into 4 positions and flipped over giving 8
positions. So the answer is
6*5*4*3/8=45 answer
Before sending, I look at Jim's answer. We have the same answer for
question 1 but wildly different answers for 2 and 3. Maybe I'm not
interpreting the problem correctly. For example, If we have a ring
of 4 colors and rotate the cube inside this ring, there are 6 sides
we can have forward and (4) 90-degree turns we can make or 24
positions. So if we count these duplicates the answer would be
45*24. I would not be able to spot duplicates from the way Jim has
converted it into 6 digits.
Regards,
Brian
----- Original Message -----
From: Brian Edward Jensen
To: Math4u@yahoogroups.com
Sent: Thursday, December 27, 2007 9:13 PM
Subject: [Math4u] Re: Cube problem
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
Answer corrected above
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
Yahoo! Groups Links
<*> To visit your group on the web, go to:
http://groups.yahoo.com/group/Math4u/
<*> Your email settings:
Individual Email | Traditional
<*> To change settings online go to:
http://groups.yahoo.com/group/Math4u/join
(Yahoo! ID required)
<*> To change settings via email:
mailto:Math4u-digest@yahoogroups.com
mailto:Math4u-fullfeatured@yahoogroups.com
<*> To unsubscribe from this group, send an email to:
Math4u-unsubscribe@yahoogroups.com
<*> Your use of Yahoo! Groups is subject to:
No comments:
Post a Comment