Question

The number of ways of arranging all the letters of the word PERFECTION such that there must be exactly two consonants between any two vowels is:

• A. \(4! + 6!\)
• B. \(3! + 6!\)
• C. \(2! \times 3! \times 6!\)
• D. \(\frac{6!}{4!}\)

Hint:

When specific spacing between vowels and consonants is required, arrange vowels first, then place consonants accordingly.

Answer 1

The Correct Option is C

Step 1: Identify vowels and consonants in PERFECTION
Vowels: \(E, E, I, O\) (4 vowels)
Consonants: \(P, R, F, C, T, N\) (6 consonants) Step 2: Condition: Exactly two consonants between any two vowels
Arrange vowels first: number of ways to arrange vowels \(= \frac{4!}{2!} = 12\) (because \(E\) repeats twice)
Place two consonants between vowels and at ends: consonants arranged in 6! ways Step 3: Arrange consonants in 6 positions with the condition
Number of ways to place consonants between vowels with exactly two consonants is \(2!\) Step 4: Total number of ways
\[ 2! \times 3! \times 6! = \text{(arrangement of consonant pairs)} \times \text{(arrangement of vowel slots)} \times \text{(arrangement of consonants)} \]