Factorial expression of $\frac{n!}{(n-r)!}$

Question

If $n \gt r,$ Then how is the following valid?

$$n \cdot (n-1) \cdot (n-2) \cdots (n-(r-1))=\frac{n!}{(n-r)!}?$$

I thought it the list of multiplications would be equal to $\frac{n!}{(n-(r-1))!}?$

Answer 1

$(n-(r-1))=(n-r+1)$, so the LHS stops immediately before the term $(n-r)$.

Answer 2

You're counting up instead of down. Note that the numbers being added to $n$ are counting down ($0, -1, -2$, etc). But instead of the number before $(n-(r-1))$ being $(n-r)$, it is $(n-(r-2))$ because $r-1 \gt r-2$.

Answer 3

$\require{cancel}$Maybe a concrete example will make things clearer: \begin{align} & \frac{9!}{(9-4)!} = \frac{9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{5\cdot4\cdot3\cdot2\cdot1} = \frac{9\cdot8\cdot7\cdot6\cdot\cancel{5\cdot4\cdot3\cdot2\cdot1}} {\cancel{5\cdot4\cdot3\cdot2\cdot1}} = 9\cdot8\cdot7\cdot6 \\[12pt] = {} & \underbrace{9\cdot8\cdot7\cdot 6}_{\text{4 factors since} \\ 9 \,-\, 4 \text{ rather} \\ \text{than } 9 \text{ minus} \\ \text{something else} \\ \text{appeared.}} \quad \text{and the last factor, 6, is } (9-4+1). \end{align}