Question:medium

In how many ways can the letters of the word ``ASSASSINATION'' be arranged so that all \(S\)'s are together?

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When some identical letters must always stay together, treat them as one block first, then arrange the remaining objects while dividing by factorials of repeated letters.
Updated On: Jun 26, 2026
  • \(10!\)
  • \(\dfrac{10!}{3!2!2!}\)
  • \(\dfrac{9!}{3!2!2!}\)
  • \(9!\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Count letter frequencies in the word ASSASSINATION.
Writing out the word: A-S-S-A-S-S-I-N-A-T-I-O-N. Count each letter: A=3, S=4, I=2, N=2, T=1, O=1. Total: $3+4+2+2+1+1=13$ letters.
Step 2: Bundle the four S letters into one block.
Since all S letters must appear together, we treat the four S letters as a single indistinguishable unit, call it [SSSS]. The problem then reduces to arranging 10 objects: [SSSS], A, A, A, I, I, N, N, T, O.
Step 3: Identify repeated objects among the 10.
Among the 10 objects: A appears 3 times, I appears 2 times, N appears 2 times. The S-block, T, and O each appear once. So we have repeated elements that would lead to duplicate arrangements if we do not divide by factorials.
Step 4: Apply the multinomial coefficient formula.
When arranging $n$ objects where groups of identical items of sizes $p, q, r, \ldots$ exist, the number of distinct arrangements is $\frac{n!}{p!\,q!\,r!\cdots}$. Here: $\frac{10!}{3!\cdot 2!\cdot 2!}$.
Step 5: Why we do not need to divide further for the S-block.
The four S letters are all identical, so within the S-block there is only one distinct internal arrangement. We already accounted for this by treating them as one indistinct unit. No additional division is needed.
Step 6: State the final answer.
Number of arrangements $= \frac{10!}{3!\cdot 2!\cdot 2!}$. \[ \boxed{\dfrac{10!}{3!\,2!\,2!}} \]
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