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In how many ways the letters of the word EDUCATION can be rearranged so that exactly 3 vowels are together?

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by chityin

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by chityin

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There are 5 vowels and 4 consonants. they are all different and unique fortunately.

the number of ways of selecting 3 vowels from 5 : 5P3 = 120/2 = 60

There are only two vowels remaining. so they can be together or they can be separate.

Let us combine the three vowels in to one single letter X.

Now we have 7 letters: X, two remaining vowels, 4 consonants.

Let us arrange the 4th vowel, 5th vowel and 4 consonants first. They can be arranged in 6! ways.

1) Out of these 6!, in 5! permutations, the two vowels are next to each other. In such arrangements, we can place X in 4 places and 3 places are not allowed. example:

C C C V4 V5 C, V4 V5 C C C C

Hence we have 5! * 4 arrangements for X.

2) Further, in the remaining (6! - 5!) arrangements, we can place X in 3 places only as we cannot put X on either side of the two vowels.

example: C C V4 C V5 C ; V4 C C C V5 C

Hence the number of arrangements allowed for X is : (6! - 5!) * 3 for X

Hence total number of all arrangements so that exactly 3 vowels occur together :

= 5! * 4 + (6! - 5!) * 3 = 5! * (4 + 15) = 19 * 5! = 2280

the number of ways of selecting 3 vowels from 5 : 5P3 = 120/2 = 60

There are only two vowels remaining. so they can be together or they can be separate.

Let us combine the three vowels in to one single letter X.

Now we have 7 letters: X, two remaining vowels, 4 consonants.

Let us arrange the 4th vowel, 5th vowel and 4 consonants first. They can be arranged in 6! ways.

1) Out of these 6!, in 5! permutations, the two vowels are next to each other. In such arrangements, we can place X in 4 places and 3 places are not allowed. example:

C C C V4 V5 C, V4 V5 C C C C

Hence we have 5! * 4 arrangements for X.

2) Further, in the remaining (6! - 5!) arrangements, we can place X in 3 places only as we cannot put X on either side of the two vowels.

example: C C V4 C V5 C ; V4 C C C V5 C

Hence the number of arrangements allowed for X is : (6! - 5!) * 3 for X

Hence total number of all arrangements so that exactly 3 vowels occur together :

= 5! * 4 + (6! - 5!) * 3 = 5! * (4 + 15) = 19 * 5! = 2280