Prove that the product of $2n$ consecutive negative integers is divisible by $(2n)!$

Asked Sep 19 '23 at 14:42

Active Sep 19 '23 at 14:42

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My approach

Let $x$ be a sufficiently large number such that the product of numbers from $x$ to $x-(2n-1)$ is $P$

$$P=x(x-1)(x-2) \ldots (x-(2n-1))$$ $$P(x-2n)!= x!$$ $$P=\frac{x!}{(x-2n)!}$$

Now, we have to prove that $(2n)!$ divides $P$.

I can't take it from here. I tried proving by expanding it but failed. Any other approach to the problem is welcome.

asked Sep 19 '23 at 14:42

PandaScientist

Thank You and yes, this answers my question – PandaScientist Sep 19 '23 at 14:48
If something is an answer to a counting problem, it must be an integer. – JMoravitz Sep 19 '23 at 14:48