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
Numerical:
Find the number of 9 letters word that can be formed using letters of ALLAHABAD.
Solution:
Let’s 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
But out of these words few words will be same as the there are repeated letters in ALLAHABAD.
A – 4 times, L 2 times. Thus we have to ignore these repeated words. This can be done by dividing the solution by 4! & 2!.
Thus number of 9 letters word formed using letters of ALLAHABAD is
=7560.
Now, let’s solve using formula.
Number of 9 letters word formed using letters of = 7560
In some cases, we need to fix some of the objects are certain positions while finding the permutation. For example if we have to arrange 5 students in 4 seats with a condition that 1st seat is always occupied by a particular Student.
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