Thanks all and Bunuel. That is the OA.

I have a doubt still..

Wont we multiply 6!5! by 2!, as there are two possibilities, one we fix women in first position and other that we fix a man in hte first position?

meghash3 wrote:

No of way in which 5 men and 6 women can be seated 1) in a row 2 ) around a table, such that no two men or women are together.

(1) In how many different ways 5 men and 6 women can be seated in row so that no 2 women or 2 men are together?

# of ways 6 women can be seated in a row is

6!. If men will be seated between them (there will be exactly 5 places between women) no 2 men or 2 women will be together. # of ways 5 men can be seated in a row is

5!. So total

6!*5!.

(2) In how many different ways 5 men and 6 women can be seated around the table so that no 2 women or 2 men are together?

Zero. 6 women around the table --> 6 places between them and only 5 men, so in any case at least 2 women will be together.

If the question were: "In how many different ways 6 men and 6 women can be seated around the table so that no 2 women or 2 men are together?"

Then: 6 men around the table can be seated in

(6-1)!=5! # of ways (# of circular permutations of

n different objects is

(n-1)!). 6 women between them can be seated in

6! # of ways. Total:

5!6!.

Hope it helps.