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: Identifying the letters
The word "PERMUTATION" consists of 11 letters: \[ P, E, R, M, U, T, A, T, I, O, N. \] The vowels are \( E, U, A, I, O \) (5 vowels), and the consonants are \( P, R, M, T, T, N \) (6 consonants).
Step 2: Arranging the consonants first
The 6 consonants (PRMTTN) are arranged, considering the repetition of T: \[ \frac{6!}{2!}. \]
Step 3: Placing vowels in available slots
Since vowels should not be adjacent, they must be placed in the 7 available gaps among the consonants. The number of ways to choose 5 out of 7 slots: \[ \binom{7}{5} = 7C_5. \]
The 5 vowels can be arranged among themselves: \[ 5!. \]
Step 4: Eliminating cases where T’s are together
Among these permutations, the number of cases where both 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 \( \boxed{(b)} \).