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AP EAMCET · Maths · Permutation Combination

If 3 sisters and 8 brothers are together playing a game, then the number of ways in which all the sisters and brothers are to be seated around a circle such that all the three sisters are not seated together is

  1. A \(8!\times 504\)
  2. B \(11!\times 8\)
  3. C \(7!\times 210\)
  4. D \(8!\times 84\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(8!\times 84\)

Step-by-step Solution

Detailed explanation

Total arrangements: \((11-1)! = 10!\) Arrangements where 3 sisters are together: \((8+1-1)! \times 3! = 8! \times 3!\) Arrangements where sisters are not together: \(10! - (8! \times 3!) = 10 \times 9 \times 8! - 6 \times 8! = 8!(90 - 6) = 8! \times 84\)