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 ⁿP_r. 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 ⁿP_r,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.