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Question: simplify $$\frac{(n-1)!}{(n-2)!}$$

What I did was:

$$\frac{(n - 1)!}{(n - 2)! \times (n - 3)!}$$ This I did following the rule $n! = n \times (n - 1)!$.

But my answer just doesn't look correct and I don't have a solution guide that tells me the correct answer. This is why I don't know whether my answer is correct or not.

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2 Answers

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Note that $(n-1)! = \color{blue}{(n-1)}(n-2)\cdots (2)(1) = \color{blue}{(n-1)}(n-2)!$ by plugging in $(n-2)! = (n-2)\cdots (2)(1)$ into the expression for $(n-1)!$.

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

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you already know that $$n!=\left( n-1 \right) !n\\ \left( n-1 \right) !=\left( n-2 \right)! \left( n-1 \right) \\ \left( n-2 \right) !=\left( n-3 \right) !\left( n-2 \right) \\ ......$$ so

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

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