If one work can be done by "a" different styles and a second work can be done by "b" different styles then the total syles of doing both works will be \((a \times b)\).
The factorial of an integer is the product of all less than or equal positive integers. It is denoted by \(n!\), where n is any positive integer. The factorial of a negative integer is not possible.$$ n! = n \ (n - 1) \ (n - 2)......3 \times 2 \times 1 $$ $$ OR $$ $$ n! = 1 \times 2 \times 3.........(n - 1) \times n $$ $$ n! = (n - 1)! \times n $$
Example: \(4! = 4 \times 3 \times 2 \times 1 = 24\)
Example: \(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1\) \(= 720\)
Note: Remember, \(0! = 1\), \(1! = 1\), \(nP_0 = 1\), \(nP_n = n!\).
The selection of objects in an ordered manner is called a permutation.
Example: Three pens A, B, and C can be written in six different orders ABC, ACB, BAC, BCA, CAB, CBA Hence here the total number of permutation are 6.
If there are "n" distinct objects and we select "r" objects together then the number of permutations is denoted as \(nP_r\) or P(n, r). $$ nP_r = n \ (n - 1) \ (n - 2)......r $$ $$ nP_r = \frac{n!}{(n - r)!} $$ Here \((r \lt n)\).
Where, n = Total number of objects.
P = Permutation.
r = The number of selected objects at a time from the total number of distinct objects.
(i): n = total number of objects.
r = Number of selected objects at a time from the total number of distict objects. $$ nP_r = \frac{n!}{(n - r)!} $$
(ii): n = total number of objects.
p, q, r = the number of different selected objects at a time from the total number of distinct objects. $$ P(p, q, r) = \frac{n!}{p! q! r!} $$
Example: Find the value of \(10P_2\)?
Solution: $$ nP_r = \frac{n!}{(n - r)!} $$ $$ 10P_2 = \frac{10!}{(10 - 2)!} $$ $$ = \frac{10!}{8!} $$ $$ = \frac{10 \times 9 \times 8!}{8!} $$ $$ 10P_2 = 90 $$