Question

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

• A. \(10!\)
• B. \(\dfrac{10!}{3!2!2!}\)
• C. \(\dfrac{9!}{3!2!2!}\)
• D. \(9!\)

Hint:

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.

Answer 1

The Correct Option is B

Step 1: Count the letters in the word.
The word is ASSASSINATION The frequency of letters is: \[ A=3,\quad S=4,\quad I=2,\quad N=2,\quad T=1,\quad O=1 \] Total letters are \[ 3+4+2+2+1+1=13 \]

Step 2: Treat all \(S\)'s as one block.
Since all \(S\)'s must be together, we treat \[ SSSS \] as one single block.
Now the objects to arrange are: \[ (SSSS), A,A,A,I,I,N,N,T,O \] So, the total number of objects is \[ 10 \]

Step 3: Account for repeated letters.
Among these \(10\) objects, the repeated objects are: \[ A=3,\quad I=2,\quad N=2 \] Therefore, the number of distinct arrangements is \[ \frac{10!}{3!2!2!} \]

Step 4: Final conclusion.
Hence, the required number of arrangements is \[ \boxed{\frac{10!}{3!2!2!}} \]