Multiple choice

Consider the word $'PERMUTATION'$. How many permutations can be made such that a word should always start with a vowel?

  1. $\dfrac{5\times 10!}{2}$
  2. $\dfrac{10!}{2!}$
  3. $5!\times 6!$

  4. none of these
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A Correct answer
Explanation

PERMUTATION has 11 letters: P, E, R, M, U, T, A, T, I, O, N. Vowels are E, U, A, I, O (5). Consonants are P, R, M, T, T, N (6). Total letters = 11. If it starts with a vowel, there are 5 choices for the first position. The remaining 10 positions can be filled in 10! / 2! ways (due to two T's). Total = 5 * 10! / 2! = 5 * 10! / 2.

AI explanation

The word PERMUTATION has 11 letters with the letter T repeated twice, and it contains 5 distinct vowels. Fixing one vowel at the start leaves 10 positions to be filled by the remaining 10 letters. The number of arrangements for these remaining letters is 10! divided by 2! to account for the repeated T, giving the total permutations as (5 * 10!) / 2.

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