How many distinct arrangements of the letters in the word “ASSASSINATION” are possible if all four S’s must be together as a single block?

Introduction / Context:We count permutations with repeated letters under a block constraint. The four S’s must appear contiguously, treated as a single super-letter.

Given Data / Assumptions:

• Word length = 13 letters.
• Letter multiplicities (excluding the S-block): A×3, I×2, N×2, T×1, O×1; S is forced as one block.

Concept / Approach:Replace the four S’s by one block “S⁴”. Then arrange this block with the remaining letters accounting for multiplicities. Use multinomial: total items = 10 (block + other letters). Count = 10! / (3! 2! 2!).

Step-by-Step Solution:Items to arrange: {S⁴-block, A, A, A, I, I, N, N, T, O} (10 items).Number of arrangements = 10! / (3! * 2! * 2!) = 3628800 / 24 = 151200.

Verification / Alternative check:Without the block constraint, total would be 13! / (4! 3! 2! 2!). Enforcing the block correctly reduces the count to 10! / (3! 2! 2!).

Why Other Options Are Wrong:10! ignores repeated letters; 3628800 is 10! (fails to divide by repeats); “14!(4!)” is not meaningful here.

Common Pitfalls:Treating the S’s as distinct or forgetting to divide by identical letters’ factorials.

Final Answer:151200