# Simplify the factorial $\frac{(n-2)!}{(n-3)!}$

What the problem looks like: $$\frac{(n-2)!}{(n-3)!}$$

• $n-2$ isn't it? Apr 4 '17 at 13:43
• Please try to typeset your question rather than linking to an image. I'll edit it for you this (first) time :-). Apr 4 '17 at 13:47

Here's an HINT : $$(n-2)!=(n-2)(n-3)!$$

Use the definition, since $k! = k\cdot (k-1) \cdot (k-2) \cdot \ldots \cdot 2 \cdot 1$, you have: $$\frac{(n-2)!}{(n-3)!} = \frac{(n-2)(n-3)(n-4)\cdots}{(n-3)(n-4)(n-5)\cdots}$$ You can cancel a lot!

Once you get the idea, note that: $$\frac{\color{blue}{(n-2)!}}{(n-3)!} = \frac{\color{blue}{(n-2)(n-3)!}}{(n-3)!} = \cdots$$

$$\frac{(n-2)!}{(n-3)!}$$

$$=\frac{(n-2)(n-2-1)!}{(n-3)!}$$

$$=\frac{(n-2)(n-3)!}{(n-3)!}$$

$$=(n-2)$$