Isn't $ n(n-1)(n-2)...(n-m+1) $ a factorial already?

Question

Let $ m \ge 1 $ and $ n \ge 1 $ be integers

Let $A$ be a set of size $m$

Let $B$ be a set of size $n$

How many one-to-one functions $f: A \rightarrow B$ are there?

skipped stuff

$$ n(n-1)(n-2)...(n-m+1) \\ = n(n-1)(n-2)...(n-m+1) \cdot \frac{(n-m)(n-m-1)...1}{(n-m)(n-m-1)...1} \\ = \frac{n(n-1)(n-2)...1}{(n-m)(n-m-1)...1} \\ = \frac{n!}{(n-m)!} $$

My question is: Isn't $ n(n-1)(n-2)...(n-m+1) $ already a factorial? Also isn't the $+1$ there to make it so you don't multiply against $0$ for product rule?

Also how does $$ n(n-1)(n-2)...(n-m+1) \cdot \frac{(n-m)(n-m-1)...1}{(n-m)(n-m-1)...1} \\ = \frac{n(n-1)(n-2)...1}{(n-m)(n-m-1)...1} \\ $$

Answer 1

No, a factorial is something like $1\cdot2\cdot3\cdot4\cdot5$. You have something like $9\cdot8\cdot7\cdot6\cdot5$. That is, a factorial has to start from $1$ and go up - yours starts from some number and goes down (and crucially, doesn't go down all the way to $1$, otherwise it would be a factorial, just written backwards).

However, you can express products like that using factorials, which is what they're doing here. For example:

$$9\cdot8\cdot7\cdot6\cdot5=9\cdot8\cdot7\cdot6\cdot5\cdot\not4\cdot\not3\cdot\not2\cdot\not1=\frac {9!} {4!}$$

The $+1$ is not really there "to avoid multiplication by $0$", it's there because... well, because the formula is incorrect without it.

Answer 2

The notation $n!$ is used to denote the product $\prod_{j=1}^n j$ and this is commonly called "$n$ factorial." Your expression is of the form $\prod_{j=n-m + 1}^n j$ and this is thus not a factorial in this sense, as it starts at $n-m+1$ and not at $1$ (except if $n=m$).

However, there is a pair of generalization of the notion of factorials called rising and falling factorials.

There one has

• $(n)_m$ a falling factorial defined as $n(n-1) \dots (n-m+1)$, and
• $(n)^m$ a rising factorial defined as $n(n+1) \dots (n+m-1)$.

The usual factorials can be recovered as the special cases $n= m$ for falling and $n=1$ for rising.

Thus, your quantity is not a factorial in the usual sense however you could conveniently express it as $(n)_m$ a falling factorial.

For your additional question:

The product before the fraction contains all integers from $n$ to $n+m -1$ the denominator contains all integers from $n-m$ to $1$ so taking them all together these are all the integers from $n$ to $1$, which is what is written.