Find the number of ways in which 6 persons out of 5 men and 5 women can be seated at a round table such that 2 men are never together 54005300 4300 6000

Label the chairs from 1 to 6 in increasing order, label the men and the women from 1 to 5. Consider three cases:

1) There is 1 man: choose a man in 5 ways, place the man at chair 1, and permute the 5 women in the remaining chairs in 5! ways. Therefore the number of ways is 5⋅5!=600.

2) There are 2 men: choose two men in (5C2) ways, place the man with the smallest label at chair 1 and the other at chair 3, 4 or 5 (3 ways), choose four women in (5C4) ways and arrange them in the remaining chairs in 4! ways. Therefore the number of ways is (5C2)⋅3⋅(5C4)⋅4!=3600.

3) There are 3 men: choose three men in (5C3) ways, place the man with the smallest label at chair 1, and the other at chair 3 and 5 (2 ways), choose three women in (5C3) ways and arrange them in the remaining chairs in 3! ways. Therefore the number of ways is (5C3)⋅2⋅(5C3)⋅3!=1200.

Hence the total number of ways is 600+3600+1200=5400.