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This was a question that I saw on the forum (asked by ABJ11):

Peter purchased a pentagonal pen for his puppy Piper. Now Peter wants to decorate the new pen for Piper, and he would like to paint each side of the pen either red, green, or blue so that each wall is a solid color. Peter can only paint at night when Piper is sleeping, and unfortunately it is too dark for him to determine which color he is painting. So for each wall, Peter randomly chooses a can of paint and paints the wall in that color. In the morning, Peter observes the resulting color scheme. What is the probability that no two adjacent walls of the pen have the same color?

 

 

And this was the answer that was posted by CalculatorUser:

Lets do it this way

 

Find the total number of ways you can paint the wall.

 

Then find the number of favorable cases over the total cases.

 

So lets have walls

 

1     (3 ways to paint it red, green or blue)

2     (3 ways)

3     (3 ways)

4     (3 ways)

5     (3 ways)

 

So 3^5 = 243 total ways you can paint it

 

Now we try to find the number of ways we can paint it that has no two adjacent walls that are same color

 

1     (3 ways to paint it red, green, or blue)

2     (2 ways to paint it, different than first one)

3     (2 ways to paint it, different than second one)

4     (2 ways to paint it, different than third one)

5     ( 1 way to paint it different than first and fourth)

 

So 3 * 2 * 2 * 2 * 1 = 24

 

HOWEVER, the above example is when the first wall is DIFFERENT than the fourth wall.

 

So we also have to add cases where they are the same.

 

1     (3 ways to paint it red, green, or blue)

2     (2 ways to paint it, different than first one)

3     (2 ways to paint it, different than second one)

4     (1 ways to paint it, same as first)

5     ( 2 ways to paint it, because the first and fourth are the same)

 

So 3 * 2 * 2 * 1 * 2 = 24

 

 

So 24 + 24 = 48 ways to paint it with no adjacent wall same

 

48/243 = 16/81

 

The answer is 16/81??????? Someone check this

 

 

I was wondering if anyone could please explain CalculatorUser's answer more. I understand that there are more ways to paint the walls than the first way that they explained:

1     (3 ways to paint it red, green, or blue)

2     (2 ways to paint it, different than first one)

3     (2 ways to paint it, different than second one)

4     (2 ways to paint it, different than third one)

5     ( 1 way to paint it different than first and fourth)

but I am confused on the second case that they explained. Why did they make the fourth wall the same as the first wall instead of a different wall (the third wall)?

 

Thanks so much in advance

 Jan 30, 2020
 #1
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The reason I didn't respond to the original question is because it's locked

 Jan 30, 2020
 #2
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I wrote a little program on my TI-84 that (I hope) has gives the number of possibilities..

Using the digits "1", "2", and "3" instead of colors, the results are:

 

12123     21213     31212

12132     21231     31231

12312     21313     31232

12313     21321     31312

12323     21323     31321

13123     23121     32121

13132     23123     32131

13212     23131     32132

13213     23213     32312

13232     23231     32321

 

I haven't come up with a neat explanation, though.

 Jan 30, 2020

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