We have $12$ similar balls we arrange them in $10$ distinct cells.

In how many arrangements will every cell have an even amount of balls (including $0$)?

I thought to use the formula:

$$ \binom{n+r-1}{r-1} $$

And to look as if the balls come in pairs.

Namely, arrange $6$ pairs of balls in $10$ different cells.

Therefore, I get:

$$ \binom{6+10-1}{10-1} = \binom{15}{9} $$

which is not the answer.

Can someone tell me why my thinking is wrong?


  • $\begingroup$ What is the given answer? I think your answer should be correct. $\endgroup$ Aug 30, 2020 at 7:08
  • $\begingroup$ @vishnuKadiri The possible answers are (this is a question in my homeworks): $a.\binom{15}{6}2^6 \ b. \frac{12!}{2^6} \cdot 10^6, \ c. \binom{15}{6}, \ d. \binom{12}{6} \cdot 6! \cdot 10^6$ None of them fit $\binom{15}{9}$, correct me if I’m wrong. $\endgroup$
    – Alon
    Aug 30, 2020 at 7:23
  • 2
    $\begingroup$ Why not? The answer is clearly C. Since $\ \binom{n}{r} = \binom{n}{n-r}$ $\endgroup$ Aug 30, 2020 at 7:53
  • $\begingroup$ Right, my fault, thank you! $\endgroup$
    – Alon
    Aug 30, 2020 at 8:16

1 Answer 1


Community wiki post so that the question can be marked as closed.

As Vishnu Kadiri indicated in the comments, your answer is correct. As Vishnu also indicated in the comments, since $$\binom{n}{k} = \binom{n}{n - k}$$ your answer $\binom{15}{9}$ is equal to the stated answer $\binom{15}{6}$ because $$\binom{15}{9} = \binom{15}{15 - 9} = \binom{15}{6}$$


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