Piling coins (piling in bins 3/4)

1) Let's start with pile with some order, and all coins identical. Put one coin in each pile and then, there is $\Big(\!\!\Big(\begin{matrix}3\\10-3\end{matrix}\Big)\!\!\Big)={9 \choose 7}=36$ ways to do that.

2) Now, assign a different value to each coin: $10!=3'628'800$ ways to do it.

3) We don't want any order for the 3 piles: divide by $3!=6$ .

Answer is $\Big(\!\!\Big(\begin{matrix}3\\10-3\end{matrix}\Big)\!\!\Big)·\dfrac{10!}{3!}=36·\dfrac{3'628'800}{6}=\boxed{21'772'800}$ .

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Notations: $\overline{C}^n_p=\Big(\!\!\Big(\begin{matrix}n\\p\end{matrix}\Big)\!\!\Big)=\Big(\begin{matrix}n+p-1\\p\end{matrix}\Big)$ is the multicombination (combination where repetitions are permitted). See stars and bars .

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Generic formula for $c$ coins and $p$ piles: $\Big(\!\!\Big(\begin{matrix}p\\c-p\end{matrix}\Big)\!\!\Big)·\dfrac{c!}{p!} = {{c-1} \choose {c-p}}·\dfrac{c!}{p!} = \dfrac{(c-1)!·c!}{(c-p)!·(p-1)!·p!}$

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Aside note: (not needed to solve problem) This is Lah number $L(10,3)=\left\lfloor\begin{array}{c}10\\3\end{array}\right\rfloor = 21'772'800$ .

We have $\left\lfloor\begin{array}{c}n\\k\end{array}\right\rfloor = \left(\begin{array}{c}n-1\\k-1\end{array}\right) \dfrac{n!}{k!} = \dfrac{n!\,(n-1)!}{(n-k)!\, k!\,(k-1)!}$