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I just figured something out while doing this question : Prove that the product of $5$ consecutive natural numbers is divisible by $120$.

I thought that $5! = 120$ and thought of this : The product of $n$ consecutive natural numbers is divisible by $n!$

Is this some pre-existing property or theorem? If yes, what is it called? I would love to learn about it in-depth.

Let me know, thanks!

Bill Dubuque
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Rajdeep Sindhu
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2 Answers2

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Here is a combinatorial argument: you might be familiar with the fact that the number of ways to choose $k$ objects (which is obviously a positive integer) is $${n \choose k} = \frac{(n + k)!}{n! k!} = \frac{1}{k!} \frac{(n+k)!}{n!}$$ But if we look at the rightmost quantity, its just the product of $k$ consecutive numbers, from $n + 1$ to $n + k$. But for ${ n \choose k}$ to be an integer, this product of $k$ consecutive numbers must be divisible by $k!$, as you claim.

paulinho
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  • Actually, I just began 11th grade Mathematics, so I'm not so familiar with Permutations and Combinations. This was just a thought that struck my mind. I also once derived the formula for Permutations $\dfrac{n!}{(n-r)!}$ and that's all, thanks for the answer by the way... – Rajdeep Sindhu Apr 16 '20 at 15:38
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One way to see it...

Let $n\ge x$ and then not only the number $n(n-1)(n-2)\cdots(n-x+1)$ is divisible by $x!$, but the number:

$${n\choose x}=\frac{n(n-1)(n-2)\cdots(n-x+1)}{x!}$$

("$n$ choose $x$") is the number of ways to pick $x$ objects out of $n$ objects without bothering about the order in which they are picked.

Proof: first try to pick them in order ($n$ ways to pick the first, $n-1$ ways to pick the next etc. - so $n(n-1)(n-2)\cdots (n-x+1)$ ways altogether), but then if you ignore the order, then those "picks" all come in groups of $x!$ where all picks in each group are just permutations of each other.