Find the number of permutation of the letter of the word REMAINS such that the vowels always occur in odd places.

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Solution

The odd places are from the set {A,E,I}, so $3! = 6$ possibilities
The even ones are {R,M,N,S}, which gives $4! = 24$ possible arrangements
So, the Required number of permutations are= $6 \times 24 = 144$

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