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.


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

  • $\begingroup$ That makes sense. I was thinking that I had to work around with the denominator. Overthought it. Thank you very much $\endgroup$ – Utsav Jul 23 '16 at 10:07
  • 2
    $\begingroup$ The general strategy with factorials is to manipulate the bigger number to display some terms with a factor of the smaller number so you can cancel away. In this case, you'd want to manipulate $(n-1)!$ in such a way that you can get a factor of $(n-2)!$ to simplify things. $\endgroup$ – Zain Patel Jul 23 '16 at 10:09
  • $\begingroup$ Oh.. ok. Will do. Thank you $\endgroup$ – Utsav Jul 23 '16 at 10:11

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

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

  • $\begingroup$ Simple and easy to understand. Thank you $\endgroup$ – Utsav Jul 23 '16 at 10:10

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