I'm trying to search this expression online, but don't know what it's called:

$$\frac{n!}{n^x(n-x)!} = \left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right) \cdots \left(1-\frac{x-1}{n}\right)$$

Thank you for your help.

  • $\begingroup$ Is $x$ an integer between $0$ and $n$? $\endgroup$ – lhf Jan 16 '18 at 0:08
  • $\begingroup$ Ahh, you changed this after I had first read it! But it is still not true! For example if n= x= 1 then the left side is $\frac{1!}{1^1(0!)}= 1$ while the right side is $(1- 1)(1)= 0$. $\endgroup$ – user247327 Jan 16 '18 at 0:12
  • $\begingroup$ This isn't quite relevant, but it is close: you could rewrite part of the expression in terms of the Pochhammer symbol, and I wonder if (generalized) hypergeometric functions by be related. $\endgroup$ – Xander Henderson Jan 16 '18 at 0:36

If $x$ is an integer between $0$ and $n$, then that equality follows direct from the definitions. It has no special name. $$ (1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{x-1}{n}) =\frac{(n-1)(n-2)\cdots(n-x+1)}{n^{x-1}} \\=\frac{n(n-1)(n-2)\cdots(n-x+1)}{n^{x}} =\frac{n(n-1)(n-2)\cdots(n-x+1)(n-x)!}{n^{x}(n-x)!} =\frac{n!}{n^x(n-x)!} $$

  • $\begingroup$ Okay, thank you for your answer. $\endgroup$ – user521846 Jan 16 '18 at 0:11

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