SOLUTION: 8. A comittee of 10 women and 9 men is looking to create a 6 person committee. a. how many ways are there of choosing this committee? 19*18*17*16*15*14=19535040

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Question 466337: 8. A comittee of 10 women and 9 men is looking to create a 6 person committee.
a. how many ways are there of choosing this committee?
19*18*17*16*15*14=19535040
b. How many ways are there of chosing 4 men and 2 women?
(i'm confused)
c. Whats the probability that it's selected with 4 men and 2 women?
(I havent a clue)
d. What is the probility that it's selected with 5 women and 1 man?
( i do not know)
Please help! Thanks.
Found 2 solutions by richard1234, stanbon:
Answer by richard1234(7193) (Show Source):
You can put this solution on YOUR website!
a: Order doesn't matter, so it is 19C6 = (19*18*...*14)/6!

b: Choose the men separately, and the women separately. Hence there are 9C4*10C2 ways.

c: The probability is simply the answer to b) over the answer to a), assuming all combinations are equally likely.

d: We do d) the same way as we did b). There are 10C5*9C1 ways to select such a committee. Divide this by the total number to obtain the probability.

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
. A comittee of 10 women and 9 men is looking to create a 6 person committee.
a. how many ways are there of choosing this committee?
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Ans: 19C6 = 19!/(13!*6!) = (19*18*...*14)/(1*2*...6) = 27,132
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b. How many ways are there of chosing 4 men and 2 women?
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Ans: [9C4*10C2] = 5670
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c. Whats the probability that it's selected with 4 men and 2 women?
Ans: 5670/27132 = 0.2090
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d. What is the probility that it's selected with 5 women and 1 man?
Ans: [10C5*9C1]/[19C5] = 0.0836
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Cheers,
Stan H.