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Question

Difficulty: HardPermutations

In how many different ways can the letters of the word SUCCESS be arranged such that the three 'S's do not all come together?

  1. 360Answer
  2. B
    420
  3. C
    60
  4. D
    35

Answer

The letters of the word SUCCESS can be arranged in 360 ways such that the three 'S's do not all come together.
The correct answer is calculated using complementary counting. First, the total unrestricted permutations of SUCCESS (7 letters with 3 'S's and 2 'C's) is 7!3!2!=420\frac{7!}{3!2!} = 420. Next, treating the three 'S's as one single block leaves 5 items to arrange with 2 'C's, giving 5!2!=60\frac{5!}{2!} = 60 ways where the 'S's are together. Subtracting 60 from 420 yields 360.

Step-by-Step Solution

1
Calculate the total number of unrestricted arrangements of the word SUCCESS.
The word SUCCESS has 7 letters in total: 3 'S's, 2 'C's, 1 'U', and 1 'E'. Total arrangements Ntotal=7!3!×2!=50406×2=420N_{total} = \frac{7!}{3! \times 2!} = \frac{5040}{6 \times 2} = 420.
Repeated letters must be accounted for by dividing the factorial of the total count by the factorials of the counts of repeated letters.
2
Calculate the number of arrangements where the three 'S's are all together.
Treat the three 'S's as a single entity (SSS). We now arrange 5 entities: (SSS), U, C, C, E. Since 'C' appears twice, Ntogether=5!2!=1202=60N_{together} = \frac{5!}{2!} = \frac{120}{2} = 60.
Grouping restricted identical items into a single block allows us to find the subset of arrangements where they stay together.
3
Subtract the number of 'together' arrangements from the total arrangements.
Nnot_together=Ntotal−Ntogether=420−60=360N_{not\_together} = N_{total} - N_{together} = 420 - 60 = 360.
The complementary counting principle gives the number of ways where the restriction is satisfied.

Key Concept

Permutations with Repeated Elements and Complementary Counting
Estimated Time:1m 30s
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Topics

Permutations

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