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In a playground, 3 sisters and 8 other girls are playing together. In a particular game, how many ways can all the girls be seated in a circular order so that the three sisters are not seated together?

(a) 457993

(b) 3386880

(c) 6544873

(d) 56549

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Asked question is from Counting in section Counting of Discrete Mathematics

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Right option is (b) 3386880

For explanation: There are 3 sisters and 8 other girls in total of 11 girls. The number of ways to arrange these 11 girls in a circular manner = (11– 1)! = 10!. These three sisters can now rearrange themselves in 3! ways. By the multiplication theorem, the number of ways so that 3 sisters always come together in the arrangement = 8! × 3!. Hence, the required number of ways in which the arrangement can take place if none of the 3 sisters is seated together: 10! – (8! × 3!) = 3628800 – (40320 * 6) = 3628800 – 241920 = 3386880.

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