The number of different words that can be for | vedclass

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Question

(A) The word $APPLICATION$ consists of $11$ letters: $A, P, P, L, I, C, A, T, I, O, N$.The vowels are $A, I, A, I, O$ ($5$ vowels) and the consonants

Vedclass · Accepted Answer

$(45)7!$

Vedclass · Answer

(A) The word $APPLICATION$ consists of $11$ letters: $A, P, P, L, I, C, A, T, I, O, N$.The vowels are $A, I, A, I, O$ ($5$ vowels) and the consonants are $P, P, L, C, T, N$ ($6$ consonants).First,arrange the $6$ consonants: $P, P, L, C, T, N$. The number of ways to arrange these is $\frac{6!}{2!} = 360$.There are $7$ possible gaps created by these $6$ consonants (including the ends) where the $5$ vowels can be placed so that no two vowels are together.The number of ways to choose $5$ gaps out of $7$ is $^7C_5 = \frac{7 	imes 6}{2} = 21$.The $5$ vowels $(A, A, I, I, O)$ can be arranged in these $5$ chosen gaps in $\frac{5!}{2!2!} = \frac{120}{4} = 30$ ways.Total number of words $= \frac{6!}{2!} 	imes ^7C_5 	imes \frac{5!}{2!2!} = 360 	imes 21 	imes 30 = 226800$.Calculating $(45)7! = 45 	imes 5040 = 226800$.Thus,the correct option is $(45)7!$.

The number of different words that can be formed using all the letters of the word $APPLICATION$ such that no two vowels come together is -

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