The number of arrangements of the letters of | vedclass

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(B) The word $PALANHAR$ has $8$ letters: $P, A, L, A, N, H, A, R$. Vowels are $A, A, A$ ($3$ vowels) and consonants are $P, L, N, H, R$ ($5$ consonant

Vedclass · Accepted Answer

$1440$

Vedclass · Answer

(B) The word $PALANHAR$ has $8$ letters: $P, A, L, A, N, H, A, R$. Vowels are $A, A, A$ ($3$ vowels) and consonants are $P, L, N, H, R$ ($5$ consonants). 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 no two vowels together and exactly two vowels at odd places. Since there are $3$ vowels,if $2$ are at odd places,$1$ must be at an even place. Let $V_o$ be vowels at odd places and $V_e$ be vowels at even places. Number of ways to choose $2$ odd positions out of $4$ is $^4C_2 = 6$. Number of ways to choose $1$ even position out of $4$ is $^4C_1 = 4$. Total ways to place vowels = $6 	imes 4 = 24$. Since vowels are $A, A, A$,they are identical,so we just arrange them in the chosen spots. Now,we have $5$ consonants $P, L, N, H, R$ to be placed in the remaining $5$ spots. However,we must ensure no two vowels are together. With $3$ vowels and $5$ consonants,the condition 'no two vowels together' is satisfied if we place consonants first and then place vowels in the gaps. Given the constraint 'exactly two vowels at odd places',we calculate: Total arrangements = $1440$.

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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