# Factorial of a Non – Negative Integer

## Factorial of a Non – Negative Integer

If n is a non-negative integer then factorial n is denoted by n! (or) n defined as follows

(i) 0! = 1

(ii) If n > 0 then n! = n (n – 1)!

The number of permutations of n dissimilar things taken all at a time is $$^{n}{{P}_{n}}=n!$$.

$$^{n}{{P}_{r}}{{=}^{(n-1)}}{{p}_{r}}{{+}^{(n-1)}}{{p}_{r-1}}$$.

Note: If n is a positive integer, then n! is the product of first n positive integers i.e., n! = 1.2.3.4.5.6….n.

Example: 9! = 1. 2. 3. 4. 5. 6. 7. 8. 9 = 362880.

Theorem: $$^{n}{{P}_{r}}=\frac{n!}{(n-r)!}$$.

Proof: $$^{n}{{P}_{r}}=n(n-1)(n-2)…(n-r+1)$$.

$$=\frac{n(n-1)(n-2)….(n-r+1)(n-r)(n-r-1)…….1}{(n-r)(n-r-1)……1}$$.

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

Example: Find the value of $$^{10}{{P}_{2}}$$.

Solution: $$^{10}{{P}_{2}}=\frac{10!}{(10-2)!}$$,

$$=\frac{10!}{8!}$$,

$$=\frac{10\times 9\times 8!}{8!}$$,

$$=10\times 9=90$$.

Example: How many four digited numbers can be formed using the digits 1, 2, 5, 7, 8, 9? How many of them begin with 9 and end with 2?

Solution: The number of four digited numbers that can be formed with the given digits = $$^{6}{{P}_{4}}=360$$.

Now the first place and the last place can be filled with 9 and 2 respectively in 1 way

The remaining two places can be remaining four digits 1, 5, 7, 8 in $$^{4}{{P}_{2}}=12$$.

The required number of ways = $$1{{\times }^{4}}{{P}_{2}}=12$$.