# In how many ways can the letters of the word accommodation be arranged if all the letters are used and all the vowels are together?

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There are 13 letters in the word ACCOMMODATION. The number of unique letters is 8, with 2 A, 2 C, 3 O and 2 M.

There are 6 vowels and 7 consonants. If the number of vowels is kept together, keeping in mind the 2 A and 3 O, the number of ways that can be done is `(6!)/(2!*3!)` = 60

Take all the vowels together as a single unit. Along with the 7 consonants there are 8 elements to be rearranged. This includes 2 M and 2 C. The number of possible arrangements here is `(8!)/(2!*2!)` = 10080.

The product of 10080 and 60 is 604800.

**There are 604800 ways in which the letters of the word ACCOMMODATION can be arranged with the vowels together.**

ACCOMMODATION is a thirteen letter word.

There are 6 vowels in it. Taking all the vowels together as a single unit, there would be 2 Cs, 2 Ms, 1D, 1 T and 1 N, i.e. 8 units altogether, including several repetitions.

There can be `(8!)/(2!*2!)` different arrangements possible with these words.

Again, among the 6 vowels there are 3 Os and 2 As. They can be arranged in `(6!)/(3!2!)` different ways.

Therefore, the number of ways the letters of the word “**accommodation” **can be arranged, if all the letters are used and all the vowels are together, is

`((8!)/(2!*2!))* ((6!)/(3!2!))= (8!*6!)/(3!*2!*2!*2!)=604800`

A-2 ,C-2, D-1,I-1,M-2,N-1, O-3 and T-1

Three vovels A ,I and O

Let consider all vovels as X=AIO

Now we have 6 letters , X , C , D , M , N , T and these letters can be arranged and form `(7!)/(2!xx2!)=1260` words

Three vowels together can be arranged and form `(6!)/(2!xx3!)`

`=60` words

By Fundamental principle of counting all letters are usedand all vowels are together form `60xx1260=75600` words