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I think we can assume every b is a box and every a is a ball, and it looks like there are 10 boxes and 6 balls. So I think there are C(15 5) (15 choose 5) ways for the combination. But the correct answer is 11 choose 5. I wonder why.

The original question is 'How many ways of arranging 6 a's and 10 b's with no consecutive a's?'

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  • $\begingroup$ Welcome to MSE. Please include your question in the body of the question, instead of putting it only in the title. $\endgroup$ – José Carlos Santos Sep 20 '18 at 8:55
  • $\begingroup$ @JoséCarlosSantos Ok, I have edited. $\endgroup$ – An Yan Sep 20 '18 at 8:59
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If you write all the $b$'s in a row

\begin{align*} b \quad b \quad b \quad b \quad b \quad b \quad b \quad b \quad b \quad b \end{align*} then you have $11$ different places to put the six $a$'s. Since you can't choose the same place twice (that would mean that two $a$'s are placed consecutively), the number of possible ways of placing the $a$'s is ${{11}\choose{6}}$. But by symmetry of the binomial coefficient, this is the same as ${{11}\choose{5}}$.

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  • $\begingroup$ Ok, I see. Thank you very much! $\endgroup$ – An Yan Sep 20 '18 at 9:11
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Since no consecutive a's, there must be at least one b in between a's: $$a_1\ b \ a_2 \ b \ a_3 \ b \ a_4 \ b \ a_5 \ b \ a_6.$$ The rest $5$ b's we must place in $7$ positions, that is: $$x_1+x_2+x_3+x_4+x_5+x_6+x_7=5, 0\le x_i\le 5.$$

Using Stars and Bars, it is: $${5+7-1\choose 7-1}={11\choose 6}={11\choose 5}.$$

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