Simplifying Expression Factorial Expression

Question

I'm confused as how I'm meant to simplify this:$$\frac{(n-2)!}{(n-2-r)!}$$

I have other factorial questions where the variable isn't present in the top factorial like the question above and I'm trying to figure out how I simplify.

Thanks

Answer 1

As noted in the comments, here is a useful hint:

If $r \geq 0$, then $$\frac{(n-2)!}{(n-2-r)!} = \frac{(n-2)(n-3) \dots (n-2-r + 1)(n-2-r)\dots (2)(1)}{(n-2-r)!}$$

If the $r$ in the expression makes it confusing, try letting $r$ equal a positive integer. Say, $r = 2$.

Answer 2

\begin{align*} \frac{(n-2)!}{(n-2-r)!} &=\frac{(n-2)!}{(n-(2+r))!} \\ &= \frac{(n-2)(n-(2+1))\dotsm(n-(2+(r-1)))(n-(2+r))!}{(n-(2+r))!}\\ &=(n-2)(n-3)\dotsm(n-1-r)\\ &=\prod_{k = 0}^{r-1}(n-(2+k))\\ &=\prod_{k=0}^{r-1}(n-2-k) \end{align*}

For example, let $r = 3$. Then \begin{align*} \frac{(n-2)!}{(n-2-3)!}&=\frac{(n-2)(n-3)(n-4)(n-5)!}{(n-5)!}\\ &=(n-2)(n-3)(n-4) \end{align*}

Using the formula $$\prod_{k = 0}^{3-1} (n-2-k) = \prod_{k=0}^2(n-2-k) = (n-2-0)(n-2-1)(n-2-2) = (n-2)(n-3)(n-4).$$