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