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 = \(1{{\times }^{4}}{{P}_{2}}=12\).