The number of arrangements of the letters of | vedclass

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(C) The word $PALANHAR$ has $8$ letters: $P, A, L, A, N, H, A, R$. There are $3$ vowels $(A, A, A)$ and $5$ consonants $(P, L, N, H, R)$. Positions ar

Vedclass · Accepted Answer

$2880$

Vedclass · Answer

(C) The word $PALANHAR$ has $8$ letters: $P, A, L, A, N, H, A, R$. There are $3$ vowels $(A, A, A)$ and $5$ consonants $(P, L, N, H, R)$. Positions are $1, 2, 3, 4, 5, 6, 7, 8$. Odd positions are $1, 3, 5, 7$ ($4$ positions) and even positions are $2, 4, 6, 8$ ($4$ positions). We need exactly $2$ vowels at odd places and $1$ vowel at an even place. Step $1$: Select $2$ odd places out of $4$ in $^4C_2 = 6$ ways. Step $2$: Select $1$ even place out of $4$ in $^4C_1 = 4$ ways. Step $3$: Arrange $3$ identical vowels $(A, A, A)$ in these $3$ chosen places in $1$ way. Step $4$: Arrange $5$ distinct consonants $(P, L, N, H, R)$ in the remaining $5$ places in $5! = 120$ ways. Condition: No two vowels are together. Total arrangements = $6 	imes 4 	imes 1 	imes 120 = 2880$.

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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