# Math Help - Combinatorics?

1. ## Combinatorics?

There are two children $C_1$ and $C_2$. $C_1$ has 12 different toys and $C_2$ has 13 different toys. Find the number of ways in which $C_1$ and $C_2$ can exchange their toys in such a way that after exchanging they still have same number of toys but not the same set.

2. $
\binom{25}{12}-1
$

Let us take the toys from both of them.

Now we select $12$ toys and give them to $C_{1}$ and the remaining $13$ to $C_{2}$ which can be done in $\binom{25}{12}$

This also contain the ways in which they have their original set of toys.

Hence,the answer is $\binom{25}{12}-1$