Permutations And Combinations Question 11

Question: In how many ways 7 men and 7 women can be seated around a round table such that no two women can sit together [EAMCET 1990; MP PET 2001; DCE 2001; UPSEAT 2002;Pb. CET 2000]

Options:

A) $ {{(7!)}^{2}} $

B) $ 7!\times 6! $

C) $ {{(6!)}^{2}} $

D) $ 7! $

Show Answer

Answer:

Correct Answer: B

Solution:

  • Fix up 1 man and the remaining 6 men can be seated in 6! ways. Now no two women are to sit together and as such the 7 women are to be arranged in seven empty seats between two consecutive men and number of arrangement will be 7!. Hence by fundamental theorem the total number of ways = 7! × 6!.



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