Permutation of n object has some of repeated kind.

Jun 16, 2020, 16:45 IST

Permutation of n object has some of repeated kind

Definition and formula description :-
The number of permutations of 'n' objects where p1 objects are of one kind, p2 objects are of one other kind… till pk, is:Permutation of n object has some of repeated kind

Application
The number of ways in which you can jumble the alphabets of the word 'balloon' is given byPermutation of n object has some of repeated kind

Example 1 :-If there are 4 chocolate chip, 2 oatmeal, and 2 double chocolate cookies in a box, in how many different orders is it possible to eat all of these cookies?

Solution:-
There are 8 cookies, so n=8. We want to eat all of the cookies; therefore, we are choosing 8 objects at a time. In this example, r=8, and we are eating a group of cookies with flavors that repeat. In the group of 8 cookies, there are 4 chocolate chip, 2 oatmeal, and 2 double chocolate cookies, so Permutation of n object has some of repeated kindPermutation of n object has some of repeated kind

Example 2 :-In how many of the distinct permutations of the letters of the word MISSISSIPPI are possible when four I’s do not come together?

Solution :-
The word MISSISSIPPI has one M, four i’s, four s’s, two P’s and a total of 11 letters. The number of all type of arrangements possible with the given alphabetsPermutation of n object has some of repeated kind

Let us first find the case when all the I’s together and so take it as one packet or unit. So now we have one M, one unit of four i’s, four s’s, two P’s and a total of 8 units.
Therefore the number of arrangements possible when all the I’s is togetherPermutation of n object has some of repeated kind

Hence, the distinct permutations of the letters of the word MISSISSIPPI when four I’s do not come together = 34650 – 840 = 33810.
Since the number count of different alpgabets is given as:
N (occurrence of alphabet ‘b’) = 1
N (occurrence of alphabet ‘a’) = 1
Also, n(occurrence of alphabet ‘l’) = 2
n(occurrence of alphabet ‘o’) = 2
n(occurrence of alphabet ‘n’) = 1

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