How many different 6 member committees can be chosen if the women and men are to be equally represented on the committee?
Anyone know how to solve this?
2006-12-21 00:18:44 · 5 answers · asked by rock n' roll 1 in Science & Mathematics ➔ Mathematics
well, it would depend if the they (the women) only can serve on one committee,,,,,, if they can be on as many as they want, and if they dont have to be paired with different men, then the options are limitless,,,,,,,,,,, if they may only be on one committee,,, and you must have 3 women per committee , then you could have a maximum of 3 committees 3 x3 =9 with one woman left over
2006-12-21 00:24:02 · answer #1 · answered by dlin333 7 · 0⤊ 1⤋
That would be 12 choose 3 time 10 choose 3.
((12 x 11 x 10)/(3 x 2 x 1)) x ((10 x 9 x 8)/(3 x 2 x 1))
I get 1,702,800
2006-12-21 00:31:37 · answer #2 · answered by Kris K 2 · 1⤊ 1⤋
There are 12C3 = 12! / (3!*9!) = 220 ways of choosing 3 men among the 12.
There are 10C3 = 10! / (3!*7!) = 120 ways of choosing 3 women among the 10.
So we have 220 * 120 = 26,400 ways of making a committee.
Thief (above me) forgot to eliminate the duplicates caused by different orders of choosing (he did permutations instead of combinations). Kris had the right formula but his/her calculator must have been partying early...
2006-12-21 01:50:00 · answer #3 · answered by Anonymous · 0⤊ 0⤋
number of ways of choosing 3 men = 12*11*10 FOR EACH of these number of ways of choosing 3 women = 10*9*8. therefore the answer is 12*11*10*10*9*8
2006-12-21 01:40:13 · answer #4 · answered by Anonymous · 0⤊ 1⤋
At maximum 5 adult adult males: (8C0)(10C7) + (8C1)(10C6) + (8C2)(10C5) + (8C3)(10C4) + (8C4)(10C3) + (8C5)(10C2) a minimum of three females (10C3)(8C4) + (10C4)(8C3) + (10C5)(8C2) + (10C6)(8C1) + (10C7)(8C0)
2016-10-15 09:03:27 · answer #5 · answered by ? 4 · 0⤊ 0⤋