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

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    $\begingroup$ $n-2$ isn't it? $\endgroup$
    – jonsno
    Apr 4 '17 at 13:43
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    $\begingroup$ Please try to typeset your question rather than linking to an image. I'll edit it for you this (first) time :-). $\endgroup$
    – StackTD
    Apr 4 '17 at 13:47
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Here's an HINT : $$(n-2)!=(n-2)(n-3)!$$

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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: $$\frac{\color{blue}{(n-2)!}}{(n-3)!} = \frac{\color{blue}{(n-2)(n-3)!}}{(n-3)!} = \cdots$$

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

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

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

$$=(n-2)$$

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