# What is the name of this expression?

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

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

• Is $x$ an integer between $0$ and $n$? – lhf Jan 16 '18 at 0:08
• 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 $(1- 1)(1)= 0$. – user247327 Jan 16 '18 at 0:12
If $x$ is an integer between $0$ and $n$, then that equality follows direct from the definitions. It has no special name. $$(1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{x-1}{n}) =\frac{(n-1)(n-2)\cdots(n-x+1)}{n^{x-1}} \\=\frac{n(n-1)(n-2)\cdots(n-x+1)}{n^{x}} =\frac{n(n-1)(n-2)\cdots(n-x+1)(n-x)!}{n^{x}(n-x)!} =\frac{n!}{n^x(n-x)!}$$