# $12$ similar balls into $10$ different cells, where each cell will have even number of balls

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?

Thanks.

• What is the given answer? I think your answer should be correct. Aug 30, 2020 at 7:08
• @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.
– Alon
Aug 30, 2020 at 7:23
• Why not? The answer is clearly C. Since $\ \binom{n}{r} = \binom{n}{n-r}$ Aug 30, 2020 at 7:53
• Right, my fault, thank you!
– Alon
Aug 30, 2020 at 8:16

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 $$\binom{15}{9} = \binom{15}{15 - 9} = \binom{15}{6}$$