Number of Permutations/Proof 1

Theorem
Let $S$ be a set of $n$ elements.

Let $r \in \N: r \le n$.

Then the number of $r$-permutations of $S$ is:
 * ${}^r P_n = \dfrac {n!} {\left({n-r}\right)!}$

When $r = n$, this becomes:
 * ${}^n P_n = \dfrac {n!} {\left({n-n}\right)!} = n!$

Using the falling factorial symbol, this can also be expressed:
 * ${}^r P_n = n^{\underline r}$

Proof
We pick the elements of $S$ in any arbitrary order.

There are $n$ elements of $S$, so there are $n$ options for the first element.

Then there are $n-1$ elements left in $S$ that we haven't picked, so there are $n-1$ options for the second element.

Then there are $n-2$ elements left, so there are $n-2$ options for the second element.

And so on, to the $r$th element of our selection: we now have $n - \left({r-1}\right)$ possible choices.

Each mapping is independent of the choices made for all the other mappings, so by the Product Rule For Counting, the total number of ordered selections from $S$:


 * $n \left({n-1}\right) \left({n-2}\right) \ldots \left({n-r+1}\right) = n \left({n-1}\right) \left({n-2}\right) \ldots \left({n-r+1}\right) \dfrac {\left({n - r}\right)!} {\left({n - r}\right)!}= \dfrac {n!} {\left({n - r}\right)!}$

This is made more rigorous in Construction of Permutations.