Question #0ef4d

1 Answer
crayon
Jan 1, 2018

See below...

Explanation:

The #8# letters can be arranged among themselves in #8!# ways.
But the I's can be arranged among itself in 3! ways and the N's in 2! ways. So the total number number of permutations after cancelling out the similar N's and I's is #(8!)/(3!*2!)# .

For the 2nd case we can fix the the N in the beginning. The among the remaining 7 letters, N occurs once and I thrice(like last case).
So number of permutations are #(7!)/(3!)# .

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