Construction of Permutations
Theorem
The ${}^n P_n$ permutations of $n$ objects can be generated algorithmically.
By Number of Permutations, that number is given by:
- ${}^n P_n = n!$
where $n!$ denotes the factorial of $n$.
This will be demonstrated to hold.
Proof 1
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
Without loss of generality, let a set $S_n$ of $n$ objects be $\set {1, 2, \ldots, n}$.
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 ${}^n P_n$ permutations on $n$ objects, there are $\paren {n + 1} \times {}^n P_n$ permutations on $n + 1$ objects.
Hence by induction, and the recursive definition of the factorial:
- ${}^n P_n = n!$
$\blacksquare$
Proof 2
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
Without loss of generality, let a set $S_n$ of $n$ objects be $\left\{{1, 2, \ldots, n}\right\}$.
Take a permutation of $S_n$:
- $a_1 \, a_2 \, a_3 \, \ldots \, a_n$
Construct the array:
- $a_1 \, a_2 \, a_3 \, \ldots \, a_n \, \dfrac 1 2, \quad a_1 \, a_2 \, a_3 \, \ldots \, a_n \, \dfrac 3 2, \quad a_1 \, a_2 \, a_3 \, \ldots \, a_n \, \dfrac 5 2, \quad \ldots, \quad a_1 \, a_2 \, a_3 \, \ldots \, a_n \, \left({n + \dfrac 1 2}\right)$
This will contain $n + 1$ objects.
Now rename the elements of each permutation using the numbers $\left\{ {1, 2, \ldots, n}\right\}$ preserving order.
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 ${}^n P_n$ permutations on $n$ objects, there are $\left({n + 1}\right) \times {}^n P_n$ permutations on $n + 1$ objects.
Hence by induction, and the recursive definition of the factorial:
- ${}^n P_n = n!$
$\blacksquare$