## § Counting with repetitions via pure binomial coefficients

- If we want to place $n$ things where $a$ of them are of kind
`a`

, $b$ are of kind `b`

, $c$of them are kind $c$. the usual formula is $n!/(a!b!c!)$. - An alternative way to count this is to think of it as first picking $a$ slots from $n$, and then picking $b$ slots from the leftover $(n - a)$ elements, and finally picking $c$ slots from $(n - a - b)$. This becomes $\binom{n}{a}\binom{n-a}{b}\binom{n - a - b}{c}$.
- This is equal to $n!/a!(n -a)! \cdot (n-a)!/n!(n - a - b)! \cdot (n - a - b)! / c!0!$, which is equal to the usual $n!/a!b!c!$ by cancelling and setting $c = n - a - b$.
- Generalization is immediate.