Q. 65.0( 2 Votes )

How many words ca

Answer :

Given, the word MUMBAI. It has 2 M’s, and total 6 letters.

We have to find a number of words that can be formed in such a way that all M’s must come together. For example- MMBAIU, AIBMMU, UMMABI are few words among them. Notice that all M letters come in the bunch (In bold letters).


A specific method is usually used for solving such type of problems. According to that, we assume the group of letters that remain together (here M, M) are assumed to be a single letter, and all other letters are as usual counted as a single letter. Now find a number of ways as usual; the number of ways of arranging r letters from a group of n letters is equals to nPr. And the final answer is then multiplied by a number of ways we can arrange the letters in that group which has to be stuck together in it (Here M, M).


Since we know, Permutation of n objects taking r at a time is nPr,and permutation of n objects taking all at a time is n!


And, we also know Permutation of n objects taking all at a time having p objects of same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.


Now,


Letters of word MUMBAI are M, U, M, B, A, I (6 letters)


Now from our method, letters are MM, U, B, A, I. (5 letters)


A total number of arrangements of 5 letters (here all distinct) is 5!


And the total number of arrangements of grouped letters (Here M, M) .


So, our final answer for arranging the letters such that all vowels stick together equals multiplication of 5! and


Total number of arrangements


= 5!


= 120


Hence, a total number of words formed during the arrangement of letters of word MUMBAI such that all M’s remains together equals to 120.


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