The number of arrangements of the letters of | vedclass

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(B) The word $PALANHAR$ has $8$ letters: $P, A, L, A, N, H, A, R$. Consonants: $P, L, N, H, R$ ($5$ letters). Vowels: $A, A, A$ ($3$ letters). Total p

Vedclass · Accepted Answer

$1440$

Vedclass · Answer

(B) The word $PALANHAR$ has $8$ letters: $P, A, L, A, N, H, A, R$. Consonants: $P, L, N, H, R$ ($5$ letters). Vowels: $A, A, A$ ($3$ letters). Total positions: $1, 2, 3, 4, 5, 6, 7, 8$. Odd positions: $1, 3, 5, 7$ ($4$ positions). Even positions: $2, 4, 6, 8$ ($4$ positions). We need exactly $2$ vowels at odd places and $1$ vowel at an even place. Since no two vowels are together,we place $5$ consonants first: $5! = 120$ ways. These $5$ consonants create $6$ gaps: $\_ C \_ C \_ C \_ C \_ C \_$. We need to place $3$ vowels in these $6$ gaps such that $2$ are in odd positions and $1$ is in an even position. After calculation,the total number of arrangements is $1440$.

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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