Multiple choice

Different words being formed by arranging the letters of the word "INTERMEDIATE". All the words obtained are written in the form of a dictionary, lf vowels & consonants occupy their original places, then the number of permutations is

  1. $\displaystyle \frac{6!}{2!}\times\frac{6!}{2!}$
  2. $\displaystyle \frac{6!}{3!}\times\frac{6!}{2!}$
  3. $\displaystyle \frac{6!\times6!}{3!2!2!}$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

INTERMEDIATE has 12 letters: 6 vowels (I, E, E, I, A, E) and 6 consonants (N, T, R, M, D, T). Vowels: I(2), E(3), A(1). Consonants: T(2), N(1), R(1), M(1), D(1). Permutations of vowels = 6! / (2!3!) = 60. Permutations of consonants = 6! / 2! = 360. Total = 60 * 360 = 21600. The expression 6!*6! / (3!2!2!) = 720*720 / 24 = 21600.

AI explanation

The word INTERMEDIATE has 12 total letters, consisting of 6 vowels (I, E, E, E, I, A) and 6 consonants (N, T, R, M, D, T). If vowels and consonants must stay in their original type positions, we calculate the arrangements for the 6 vowel spots and 6 consonant spots separately. The 6 vowels can be arranged in 6! / (3! 2!) ways because E is repeated 3 times and I is repeated 2 times. The 6 consonants can be arranged in 6! / 2! ways because T is repeated 2 times. Multiplying these independent arrangements gives (6! times 6!) / (3! 2! 2!).

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