Solution : let ’ s solve using logical method . Since 3 digit numbers has to be formed , lets assume 3 buckets
In this case , bucket 1 can have any of the 5 digits ( 1,3,5,7,9 ). That is 5 choices
Bucket 2 can also have any of the 5 digits , since repetition is allowed , that is 5 choices
Bucket 3 can also have any of the 4 digits , since repetition is allowed , that is 5 choices
So , total number of 3 digit numbers with repetition is 5 * 5 * 5 = 125 Lets solve using formula . Total number of 3 digit numbers with repetition is n r , here n = 5 & r = 3 So , Total number of 3 digit numbers with repetition 5 3 = 125
Permutation when all objects are not distinct
Sometimes we come up with scenarios where the objects are no distinct . For example , Let ’ s take few green , blue , red & grey books . Now we have to find the number of ways we can arrange them .
Theorem 3 : The number of permutations of n objects , where p1 objects are of one kind , p2 are of second kind , ..., pk are of kth kind and the rest , if any , are of different kind is n ! /( p1 ! * p2 ! * p3 ! * … pk !)
Numerical : Find the number of 9 letters word that can be formed using letters of ALLAHABAD .
Solution : lets solve logically first . Assume the repeated words to be different . Let ALLAHABAD be
A1L1L2A2HA3BA4D
Assume 9 buckets & each of these buckets has 9,8,7,6,5,4,3,2,1 choice . So total number of 9 letters word formed is 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 362880