Case 3: Without repetition & Permutation of sub set with r elements Permutation = n *( n-1) *( n-2)…( n-r + 1) Case 4: With repetition & Permutation of sub set with r elements Permutation = n * n * n … r times
Symbols used in Permutation n! = n( n-1)( n-2) ……………….. 3.2.1.
n Pr = n!/( n-r)!
= n *( n-1) *( n-2)…( n-r + 1) For n > r Numerical: Find( i) 5!( ii) 7!/ 5!( iii) 8 P5
Solution: 5! = 5 * 4 * 3 * 2 * 1 = 120 7!/ 5! =( 7 * 6 * 5 * 4 * 3 * 2 * 1) /( 5 * 4 * 3 * 2 * 1) = 42
8 P5 = 8! /( 8-5)! = 8! / 3! =( 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / 3 * 2 * 1 = 6720
Numerical: How many ways 3 people can stand in a queue?
Solution: In this case it is implicit that repetition is not allowed, as we can’ t repeat a person.
Theorem 1: The number of permutations of n different objects taken r at a time, where 0 < r ≤ n and the objects do not repeat is n Pr. Or n( n – 1)( n – 2)...( n – r + 1) Or n!/( n-r)!
Numerical: 3 vacant seats, 6 people standing, how many ways they can sit. Solution: Lets solve this conventional way.