Question

The number of different permutations of all the letters of the word "PERMUTATION" such that any two consecutive letters in the arrangement are neither both vowels nor both identical is:

• A. \( 63 \times 6! \times 5! \)
• B. \( 57 \times 5! \times 5! \)
• C. \( 33 \times 6! \times 5! \)
• D. \( 7 \times 7! \times 5! \)

Hint:

When solving word permutation problems with restrictions, always first arrange the unconstrained group, then insert the constrained group into available slots, and finally subtract cases violating restrictions.

Answer 1

The Correct Option is B

Step 1: Identify letters
The word "PERMUTATION" has 11 letters: P, E, R, M, U, T, A, T, I, O, N. There are 5 vowels (E, U, A, I, O) and 6 consonants (P, R, M, T, T, N).
Step 2: Arrange consonants
The 6 consonants are arranged, accounting for the repeated 'T': \[\frac{6!}{2!}\]
Step 3: Place vowels in gaps
To ensure no vowels are adjacent, place them in the 7 available slots among the consonants. The number of ways to select 5 slots from 7 is \[\binom{7}{5} = 7C_5\]. The 5 vowels can then be arranged in \[\;5!\] ways.
Step 4: Exclude cases with adjacent T’s
Calculate permutations where the two 'T's are together: \[\;5! \times 6C_5 \times 5!\].
Step 5: Final Calculation \[\frac{6!}{2!} \times 7C_5 \times 5! - 5! \times 6C_5 \times 5! = 57 \times (5!)^2.\]Final Answer: The correct answer is \( \boxed{(b)} \).