Q.In how many ways can the letters of the word PERMUTATIONS be arranged if the

Concept: Permutations Without Repetition (all letters are distinct except T appears twice).

The word PERMUTATIONS has 12 letters: P, E, R, M, U, T, A, T, I, O, N, S.

T appears twice; all other letters are distinct.

1. Words start with P and end with S Fix P at the first position and S at the last. The remaining 10 letters (including two T’s) can be arranged in the middle. Number of ways = $\frac{10!}{2!}$.
2. Vowels are all together Vowels: E, U, A, I, O — 5 distinct vowels. Treat them as a single block. Now we have: block + 7 consonants (P, R, M, T, T, N, S) = 8 items, with T repeated twice. Arrange these 8 items: $\frac{8!}{2!}$. Inside the block, arrange 5 vowels: $5!$. Total = $\frac{8!}{2!} \times 5!$.
3. Always 4 letters between P and S …

Let’s first understand the word: PERMUTATIONS has 12 letters. The letters are: P, E, R, M, U, T, A, T, I, O, N, S. Notice that T appears twice — all other letters are distinct. This repetition will matter in every part.

The core idea in all three parts is permutations without repetition (except for the repeated T). When we arrange distinct objects, the number of ways is $n!$. When there are identical objects, we divide by the factorial of the count of each identical group. Here, only T is repeated, so the total unrestricted arrangements would be $\frac{12!}{2!}$.

Now, each part imposes a different condition. We’ll handle them one by one.

Part (i): Words start with P and end with S

1. Fix the first and last positions.

   If the word must start with P and end with S, then P and S are locked in place. That leaves 10 remaining positions (positions 2 through 11) to fill with the remaining 10 letters.

2. What are the remaining letters?

   After removing one P and one S, we have: E, R, M, U, T, A, T, I, O, N. That’s 10 letters, with T appearing twice.

3. Count the arrangements of these 10 letters.

   The number of distinct permutations of these 10 letters, accounting for the two identical T’s, is $\frac{10!}{2!}$.

4. No further restrictions.

   Since P and S are fixed, every arrangement of the middle 10 letters gives a valid word. So the total number is simply $\frac{10!}{2!}$.

Part (ii): Vowels are all together

1. Identify the vowels.

   In PERMUTATIONS, the vowels are: E, U, A, I, O. That’s 5 distinct vowels.

2. Treat the vowel group as a single block.

   If all vowels must stay together, we can think of them as one “super-letter”. So now we have:

   • The block of vowels (call it V)
   • The consonants: P, R, M, T, T, N, S That’s 1 block + 7 consonants = 8 items in total.

3. But note: the consonants include two T’s.

   So among these 8 items, the two T’s are identical. The number of distinct arrangements of these 8 items is $\frac{8!}{2!}$.

4. Now arrange the vowels inside the block.

   The 5 vowels are all distinct, so they can be arranged among themselves in $5!$ ways.

5. Multiply the two counts.

   Total arrangements = (arrangements of block + consonants) × (arrangements inside block)

Part (iii): There are always exactly 4 letters between P and S

1. Interpret the condition.

   “4 letters between P and S” means that in the arrangement, the positions of P and S differ by exactly 5. For example, if P is at position $i$, then S must be at position $i+5$ (or vice versa). So the pair (P, S) can appear in two orders: P _ _ _ _ S or S _ _ _ _ P.

2. First, decide the positions for P and S. …