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 .