Construction of Permutations

Theorem
The number of permutations $${}^nP_n$$ of $$n$$ objects is $$n!$$.

Proof
The following is an inductive method of creating all the permutations of $$n$$ objects.

Base Case
There is clearly one way to arrange one object in order.

Inductive Hypothesis
We assume that we have constructed all $$n!$$ permutations of $$n$$ objects.

Induction Step
WLOG let a set $$S_n$$ of $$n$$ objects be $$\left\{{1, 2, \ldots, n}\right\}$$.

Take a permutation of $$S_n$$:
 * $$a_1 \, a_2 \, a_3 \, \ldots \, a_n$$

Now we take the number $$n+1$$.

We can form $$n+1$$ permutations from this one by putting $$n+1$$ in all places possible:
 * $$a_{n+1} \, a_1 \, a_2 \, a_3 \, \ldots \, a_n, \quad a_1 \, a_{n+1} \, a_2 \, a_3 \, \ldots \, a_n, \quad a_1 \, a_2 \, a_{n+1} \, a_3 \, \ldots \, a_n, \quad \ldots, \quad a_1 \, a_2 \, a_3 \, \ldots \, a_n \, a_{n+1}$$

It is clear that all permutations of $$n+1$$ objects can be obtained in this manner, and no permutation is obtained more than once.

As there are $${}^nP_n$$ permutations on $$n$$ objects, there are $$\left({n + 1}\right) {}^nP_n$$ permutations on $$n+1$$ objects.

Hence by induction, and the recursive definition of the factorial:
 * $${}^n P_n = n!$$