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Worked Examples · Example 27

Q.In how many ways can the letters of the word ASSASSINATION be arranged? In how many of these arrangements the four S's do not come together.

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Compute total permutations with repeated letters, then subtract the "all four S's together" (block) count to get "S's not together".

[!FORMULA]

Permutations of nn letters with repetitions p1,p2,…p_1,p_2,\dots each =n!p1! p2!⋯= \dfrac{n!}{p_1!\,p_2!\cdots}. Treating kk identical repeated letters as forced-together reduces them to a single block: arrange (n−k+1)(n-k+1) units (the block among the rest), with remaining repeats still divided out; the block itself has only 11 internal arrangement since all its letters are identical.

  1. Letters of ASSASSINATION: A,S,S,A,S,S,I,N,A,T,I,O,NA,S,S,A,S,S,I,N,A,T,I,O,N — total n=13n=13 letters. Counting occurrences: A=3A=3, S=4S=4, I=2I=2, N=2N=2, T=1T=1, O=1O=1 (check: 3+4+2+2+1+1=133+4+2+2+1+1=13 ✓).
  2. Total arrangements =13!3! 4! 2! 2! 1! 1!=62270208006×24×2×2=6227020800576=10810800= \dfrac{13!}{3!\,4!\,2!\,2!\,1!\,1!} = \dfrac{6227020800}{6\times24\times2\times2} = \dfrac{6227020800}{576} = 10810800. …

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