Number of Permutations

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:
 * $${}^r P_n = \frac {n!} {\left({n-r}\right)!}$$

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

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

Informal 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 $$S$$:


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

This is made more rigorous in Construction of Permutations.

Formal Proof
From the definition, an $r$-permutations of $S$ is an ordered selection of $$r$$ elements of $$S$$.

It can be seen that an $$r$$-permutation is an injection from a subset of $$S$$ into $$S$$.

From Cardinality of Set of Injections‎, we see that the number of $$r$$-permutations $${}^r P_n$$ on a set of $$n$$ elements is given by:
 * $${}^r P_n = \frac {n!} {\left({n-r}\right)!}$$

From this definition, it can be seen that a bijection $$f: S \to S$$ is an $n$-permutation.

Hence the number of $$n$$-permutations on a set of $$n$$ elements is $${}^n P_n = \frac {n!} {\left({n-n}\right)!} = n!$$.