# Divisibility of Product of Consecutive Integers

## Theorem

The product of $n$ consecutive positive integers is divisible by the product of the first $n$ consecutive positive integers.

That is:

$\displaystyle \forall m, n \in \Z_{>0}: \exists r \in \Z: \prod_{k \mathop = 1}^n \paren {m + k} = r \prod_{k \mathop = 1}^n k$

## Proof

 $\displaystyle \prod_{k \mathop = 1}^n \paren {m + k}$ $=$ $\displaystyle \frac {\paren {m + n}!} {m!}$ $\displaystyle$ $=$ $\displaystyle n! \frac {\paren {m + n}!} {m! \, n!}$ $\displaystyle$ $=$ $\displaystyle n! \binom {m + n} m$ Definition of Binomial Coefficient $\displaystyle$ $=$ $\displaystyle \binom {m + n} m \prod_{k \mathop = 1}^n k$

Hence the result, and note that for a bonus we have identified exactly what the divisor is:

$\dbinom {m + n} m$

$\blacksquare$

## Examples

### Product of $5$ to $8$

 $\displaystyle 5 \times 6 \times 7 \times 8$ $=$ $\displaystyle 1680$ $\displaystyle$ $=$ $\displaystyle 70 \times 24$ $\displaystyle$ $=$ $\displaystyle \binom 8 4 \times 1 \times 2 \times 3 \times 4$

### Product of $10$ to $13$

 $\displaystyle 10 \times 11 \times 12 \times 13$ $=$ $\displaystyle 17 \, 160$ $\displaystyle$ $=$ $\displaystyle 715 \times 24$ $\displaystyle$ $=$ $\displaystyle \binom {13} 4 \times 1 \times 2 \times 3 \times 4$

### Product of $4$ to $8$

 $\displaystyle 4 \times 5 \times 6 \times 7 \times 8$ $=$ $\displaystyle 6720$ $\displaystyle$ $=$ $\displaystyle 56 \times 120$ $\displaystyle$ $=$ $\displaystyle \binom 8 5 \times 1 \times 2 \times 3 \times 4 \times 5$

### Product of $11$ to $15$

 $\displaystyle 11 \times 12 \times 13 \times 14 \times 15$ $=$ $\displaystyle 360 \, 360$ $\displaystyle$ $=$ $\displaystyle 3003 \times 120$ $\displaystyle$ $=$ $\displaystyle \binom {15} 5 \times 11 \times 12 \times 13 \times 14 \times 15$