Divisibility of Product of Consecutive Integers
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Theorem
The product of $n$ consecutive positive integers is divisible by the product of the first $n$ consecutive positive integers.
That is:
- $\ds \forall m, n \in \Z_{>0}: \exists r \in \Z: \prod_{k \mathop = 1}^n \paren {m + k} = r \prod_{k \mathop = 1}^n k$
Proof
\(\ds \prod_{k \mathop = 1}^n \paren {m + k}\) | \(=\) | \(\ds \frac {\paren {m + n}!} {m!}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n! \frac {\paren {m + n}!} {m! \, n!}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n! \binom {m + n} m\) | Definition of Binomial Coefficient | |||||||||||
\(\ds \) | \(=\) | \(\ds \binom {m + n} m \prod_{k \mathop = 1}^n k\) |
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 $8$
\(\ds 5 \times 6 \times 7 \times 8\) | \(=\) | \(\ds 1680\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 70 \times 24\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \binom 8 4 \times 1 \times 2 \times 3 \times 4\) |
Product of $10$ to $13$
\(\ds 10 \times 11 \times 12 \times 13\) | \(=\) | \(\ds 17 \, 160\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 715 \times 24\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \binom {13} 4 \times 1 \times 2 \times 3 \times 4\) |
Product of $4$ to $8$
\(\ds 4 \times 5 \times 6 \times 7 \times 8\) | \(=\) | \(\ds 6720\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 56 \times 120\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \binom 8 5 \times 1 \times 2 \times 3 \times 4 \times 5\) |
Product of $11$ to $15$
\(\ds 11 \times 12 \times 13 \times 14 \times 15\) | \(=\) | \(\ds 360 \, 360\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 3003 \times 120\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \binom {15} 5 \times 11 \times 12 \times 13 \times 14 \times 15\) |
Sources
- 1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {3-1}$ Permutations and Combinations: Theorem $\text {3-3}$
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $24$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $24$