# Permutations

## Permutations

Number of permutations of n different thing taken all at a time is n!

Proof:

By the multiplication rule, the number of ways of filling up the first, second, third … nth places together is n (n – 1) (n – 2) …. 2 x 1 = n!.

Thus $$^{n}{{P}_{n}}=n!$$.

$$^{n}{{P}_{n}}{{=}^{n}}{{P}_{n-1}}=n!$$.

Factorial Zero: From the formula, we have

$$^{n}{{P}_{r}}=\frac{n!}{0!}$$ … (1)

Also, the number of permutations of n different things taken all at a time in n! … (2)

From equation (1), the number of permutations of n different things taken all at a time

$$^{n}{{P}_{n}}=\frac{n!}{0!}$$ … (3)

From (2) and (3), we have

$$n!=\frac{n!}{0!}$$ … (4)

Again equation (4) will be valid only 0! Is taken as 1.

Thus 0! has no meaning from the definition of factorial

But in order to make the formula for $$^{n}{{P}_{r}}=\frac{n!}{(n-r)!}$$ valid for r = n, 0! is taken as 1.

Meaning of 1/(-k)! where k is a positive integer:

$$^{n}{{p}_{r}}=\frac{n!}{(n-r)!}$$.

Putting r = n + k, we have

$$^{n}{{p}_{n+k}}=\frac{n!}{(-k)!}$$.

But, the number of ways of arranging n + k out of n different things is 0. Therefore,

$$\frac{n!}{(-k)!}=0$$.

I.e., $$\frac{1}{(-k)!}=0$$.

Example:  If  $$^{9}{{P}_{5}}+5{{\ }^{9}}{{P}_{4}}{{=}^{10}}{{P}_{r}}$$, find the value of r.

Solution: Given that $$^{9}{{P}_{5}}+5{{\ }^{9}}{{P}_{4}}{{=}^{10}}{{P}_{r}}$$.

$$=\frac{9!}{(9-5)!}+5\ \frac{9!}{(9-4)!}$$.

$$=\frac{9!}{4!}+5\times \frac{9!}{5!}$$.

$$=\frac{9!}{4!}+5\times \frac{9!}{5\times 4!}$$.

$$=\frac{9!}{4!}+\frac{9!}{4!}$$.

$$=2\times \frac{9!}{4!}$$.

$$=2\times \frac{5\times 9!}{5\times 4!}$$.

$$=\frac{10\times 9!}{5!}$$.

$$=\frac{10!}{5!}$$.

$$^{10}{{P}_{r}}{{=}^{10}}{{P}_{5}}$$.

The value of r is 5.