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In how many ways can 10 identical red balls, 5 identical green balls and 5 identical blue balls be arranged in a row, if there can't be two adjacent balls of the same color?

What I tried: The arrangement must be of the form:

red - another color - red - another color - ... etc.

or

another color - red - another color - red ... etc.

So the problem is equivalent to find the number of arrangements of 5 green balls and 5 blue balls and then multiply by 2.

the number of arrangements of 5 green balls and 5 blue balls is $\frac{10!}{5!5!}=252$.

So, there are $2 \times 252 = 504$ arrengements.

I'd like to know if this solution is correct. Thank you for your help.

Twnk
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2 Answers2

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Consider the reds separated by $5$ blues and $4$ greens, with one green spare

$\color{red}{\Large\bullet}\color{blue}{\Large\bullet}\color{red}{\Large\bullet}\color{blue}{\Large\bullet}\color{red}{\Large\bullet}\color{blue}{\Large\bullet}\color{red}{\Large\bullet}\color{blue}{\Large\bullet}\color{red}{\Large\bullet}\color{blue}{\Large\bullet}\color{red}{\Large\bullet}\color{green}{\Large\bullet}\color{red}{\Large\bullet}\color{green}{\Large\bullet}\color{red}{\Large\bullet}\color{green}{\Large\bullet}\color{red}{\Large\bullet}\color{green}{\Large\bullet}\color{red}{\Large\bullet}$

The spare green has $7$ places to fit: $5$ right(say) of a blue, and $2$ ends, thus together with an interchange between blue/green,

Number of ways = $\boxed {2\times7\binom94 = 1764}$

true blue anil
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  • Concise and clear $+1$ – Not a Salmon Fish Apr 18 '21 at 16:07
  • Glad you like it, concise & clear is always what I aim at ! – true blue anil Apr 18 '21 at 16:59
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    If you have $5$ blue balls in place, there are $10$ places adjacent to a blue. I am guessing you had something more precise in mind than what you wrote. – N. F. Taussig Apr 19 '21 at 21:47
  • @N.F.Taussig: You are right, I have amended the wording. Actually the interchange between blue/green takes care of adjacency. – true blue anil Apr 20 '21 at 06:10
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    Your corrected answer is elegant. – N. F. Taussig Apr 20 '21 at 09:04
  • Glad you like it ! – true blue anil Apr 20 '21 at 11:23
  • @trueblueanil i did not understand something, there are $5$ place to set spare green adjacent to blue and they can interchange so $5 \times 2 \times C(9,4)$ or the spare green can be placed in the ends such that $2 \times C(9,4)$, so isnt it $12 \times C(9,4)$ –  Apr 30 '21 at 08:07
  • @leonard: When the blues are placed "inside", the spare green can be placed to the right (say) of the blues + $2$ ends in $7$ ways. Interchanging the blues and greens will place the spare blue to the right of the greens. – true blue anil Apr 30 '21 at 11:28
  • @N.F.Taussig: I happened to revisit this question, and noticed your remark re "corrected answer...."* which gives the impression that the earlier answer was incorrect. Actually, the earlier answer was also correct. If you see the edit history, it used the concept of 'mirror image' which actually I think is more elegant, but could possibly create doubts, so I just made it more explicit for easier understanding. Regards. – true blue anil Jul 09 '21 at 07:43
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There is another possibility you have not considered. There could be both a green and a blue ball between two red balls.

Line up the ten red balls. We must fill the nine spaces between successive red balls with at least one ball of another color.

If there is exactly one ball of another color between each red ball, then red balls must alternate with balls of another color. Choosing whether such an arrangement begins or ends with a red ball determines which spaces are occupied by red balls. In each case, choosing which five of the ten spaces reserved for balls of another color are occupied by the blue balls determines the arrangement. Hence, as you found, there are $$2\binom{10}{5}$$ such arrangements.

The other possibility is that one blue and one green ball are placed in the same space between successive red balls. There are nine ways to choose this space and two ways to arrange the blue and green balls within this space. That leaves eight spaces to fill with the remaining four blue and four green balls. Choose which four of those eight spaces are filled with blue balls. There are $$\binom{9}{1}2!\binom{8}{4}$$ such arrangements.

That gives a total of $$2\binom{10}{5} + \binom{9}{1}2!\binom{8}{4} = 504 + 1260 = 1764$$ admissible arrangements.

N. F. Taussig
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