Answer :

Given, the word UNIVERSITY. It has 4 vowels and 6 consonants. Vowels included in given word are (U, I, E, I) and consonants included in given word are (N, V, R, S, T, Y). In given word, one vowel letter I is repeated twice.

We have to find a number of words that can be formed in such a way that all vowels must come together. For example- NUIEIVRSTY, UIIENRSYTV, YTSRIIEUVN, and RTSEIUIYNV are few words among them. Notice that all vowel 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 U, I, E, I) 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 ^{n}P_{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 U, I, E, I).

Since we know, Permutation of n objects taking r at a time is ^{n}P_{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 the 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 UNIVERSITY are U, N, I, V, E, R, S, I, T, Y. (10 letters)

Now from our method, letters are UIEI, N, V, R, S, T, Y. (7 letters)

A total number of arrangements of 7 letters (here all distinct) is 7!

And the total number of arrangements of grouped letters (Here U, I, E, I) is .

So, our final answer for arranging the letters such that all vowels stick together equals multiplication of 7! and

Total number of arrangements

= 60480

Hence, a total number of words formed during the arrangement of letters of word UNIVERSITY such that all vowels remain together is equals to 60480.

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