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In how many ways can the letters of the word accommodation be arranged if all the...

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lilian-0716 | Student | (Level 1) Salutatorian

Posted September 12, 2013 at 9:59 AM via web

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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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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted September 12, 2013 at 1:45 PM (Answer #3)

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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.

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llltkl | College Teacher | (Level 3) Valedictorian

Posted September 12, 2013 at 12:20 PM (Answer #2)

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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` 

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aruv | High School Teacher | (Level 2) Valedictorian

Posted September 12, 2013 at 10:23 AM (Answer #1)

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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

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lxsptter | Student, Undergraduate | (Level 2) Valedictorian

Posted September 12, 2013 at 1:16 PM (Reply #1)

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Sorry for the typo in the response.

The word ACCOMMODATION has 13 letters, this includes 2 C, 2 M, 3 O and 2A. The total number of distinct letters is 8.

There are 3 vowels, A, I and O. These can be kept together in 3! = 6 different ways.

If the vowels are kept together in a unit, there are 8 - 3 + 1 = 6 different elements to be rearranged. This can be done in 6! different ways. 6!*3! = 4320

The total number of ways in which the letters of the word  ACCOMMODATION can be arranged if the vowels are together is 4320.

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