1
$\begingroup$

Over at this link, there is a claim that $(2n)! = n!n! {{2n} \choose {n}}$ - see Tom Boardman's answer, the second one down.

I'm wondering why this is the case and if anyone can provide a proof. Is this a special case of a more general relation, and does anyone have any good references for manipulating binomial coefficients and factorials?

$\endgroup$
4
$\begingroup$

For integers $a\ge b\ge 0$, we have ${a\choose b}=\frac{a!}{b!(a-b)!}$ (see wiki). Take the special case $a=2b$ and you get ${2b\choose b}=\frac{(2b)!}{b!b!}$, then cross-multiply.

$\endgroup$
1
  • $\begingroup$ That's perfect, I made a typo in my original question - glad you understood what I was after! $\endgroup$ – analystic Apr 13 '15 at 5:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.