Suppose that a department contains 10 men and 15 women. how many ways are there to form a committee with 6 members if it must have more women than men

96460 possible committees. If the 6 member committee must have more women than men, it's easier to simply enumerate the possible committees types 4 women, 2 men 5 women, 1 man 6 women, 0 men For the 4 women solution the number of ways of selecting 4 women out of a pool of 15 is 15!/(4!11!) = 1365 and then multiply by the number of ways of selecting 2 men out of a pool of 10, which is 10!/(2!8!) = 45. So we have 1365*45 = 61425 possible 4 women, 2 man committees. For the 5 women solution, the number of combinations of selecting 5 women is 15!/(5!10!) = 3003, and the number of men would be 10!/(1!9!) = 10. So there are 10*3003 = 30030 possible 5 women, 1 man committees. And finally, for the 6 woman solution, the number of combinations is 15!/(6!9!) = 5005. So the total number of possible committees is 61425 + 30030 + 5005 = 96460