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