How to simplify : $\frac{(x-4)!}{(x-1)!}$ ??

Question

Isn't $(x-4)!$ equal to $(x-4)(x-3)(x-2)(x-1)$? See the picture. How it is equating?

Thanks!

Answer 1

The thing about a factorial is that it has to go all of the way back to $1$ in the product. Thus, $(x-4)!$ factorial goes far beyond $(x-1)$, and we see the following if we expand the factorials:

$$\frac{(x-4)!}{(x-1)!} = \frac{\color{blue}{(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)\cdots(2)(1)}}{(x-1)(x-2)(x-3)\color{blue}{(x-4)(x-5)(x-6)\cdots(2)(1)}}$$

The terms in blue all cancel (and hopefully this is obvious by the expansion above), yielding the result

$$\frac{(x-4)!}{(x-1)!} = \frac{1}{(x-1)(x-2)(x-3)}$$

(Footnote: $x$ is assumed a positive integer greater than or equal to $4$ in the above algebra. Since you're taking the limit $x\to \infty$, this can be justified.)

Answer 2

$n!$ is the product of all natural numbers up to and including $n$. So

$(x-4)!=(x-4)(x-5)(x-6)\ldots 3\cdot2\cdot1$ and

$(x-1)!=(x-1)(x-2)(x-3)(x-4)(x-5)\ldots3\cdot2\cdot1$. Hence dividing gives $$\frac{(x-4)!}{(x-1)!}=\frac{1}{(x-1)(x-2)(x-3)}$$