The number of arrangements of the letters of | vedclass

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(C) The word $PALANHAR$ has $8$ letters: $P, A, L, A, N, H, A, R$. The vowels are $A, A, A$ and consonants are $P, L, N, H, R$. There are $5$ consonan

Vedclass · Accepted Answer

$2880$

Vedclass · Answer

(C) The word $PALANHAR$ has $8$ letters: $P, A, L, A, N, H, A, R$. The vowels are $A, A, A$ and consonants are $P, L, N, H, R$. There are $5$ consonants and $3$ vowels. Total positions are $8$ $(1, 2, 3, 4, 5, 6, 7, 8)$. Odd positions are $1, 3, 5, 7$ ($4$ positions) and even positions are $2, 4, 6, 8$ ($4$ positions). We need exactly two vowels at odd places and one vowel at an even place,such that no two vowels are together. First,arrange the $5$ consonants: $P, L, N, H, R$. They create $6$ gaps: $\_ C \_ C \_ C \_ C \_ C \_$. Number of ways to arrange consonants is $5! = 120$. We have $4$ odd positions and $4$ even positions. To have no two vowels together,we must place vowels in the gaps between consonants. After placing $5$ consonants,there are $6$ gaps. $3$ gaps are at odd positions and $3$ gaps are at even positions. We need $2$ vowels in odd positions and $1$ vowel in an even position. Number of ways to choose $2$ odd gaps from $4$ is $\binom{4}{2} = 6$. Number of ways to choose $1$ even gap from $4$ is $\binom{4}{1} = 4$. Total ways = $5! 	imes \binom{4}{2} 	imes \binom{4}{1} = 120 	imes 6 	imes 4 = 2880$.

The number of arrangements of the letters of the word $PALANHAR$ in which no two vowels are together and exactly two vowels are at odd places,is

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