Number of different words that can be formed | vedclass

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Question

(A) The word $APPLICATION$ has $11$ 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,

Vedclass · Accepted Answer

$(45)7!$

Vedclass · Answer

(A) The word $APPLICATION$ has $11$ 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).To ensure no two vowels come together,we first arrange the $6$ consonants: $P, P, L, C, T, N$.The number of ways to arrange these consonants is $\frac{6!}{2!}$ (since $P$ repeats twice).These $6$ consonants create $7$ gaps (including ends): $\_ C \_ C \_ C \_ C \_ C \_ C \_$.We need to place $5$ vowels in these $7$ gaps. The number of ways to choose and arrange the vowels in these gaps is $^7P_5 = \frac{7!}{2!} = 7 	imes 6 	imes 5 	imes 4 	imes 3 = 2520$.However,the vowels are $A, A, I, I, O$. The number of ways to arrange these $5$ vowels is $\frac{5!}{2!2!} = \frac{120}{4} = 30$.Total ways = $\frac{6!}{2!} 	imes ^7C_5 	imes \frac{5!}{2!2!} = 360 	imes 21 	imes 30 = 226800$.Calculating $(45)7! = 45 	imes 5040 = 226800$.Thus,the correct option is $(45)7!$.

Number of different words that can be formed from all letters of the word $APPLICATION$ such that two vowels never come together is -

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