SOLUTION: A group of 10 women and 12 men must select a 3-person committee. How many committees are possible if there must be a majority of men?

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Question 967134: A group of 10 women and 12 men must select a 3-person committee. How many committees are possible if there must be a majority of men?
Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
A majority means more than 50%

Since 1/3 = 33% roughly and 2/3 = 67% roughly, this means that if we must have a "majority of men", then the committee must have 2 or more men.

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If we have 2 men, then...

12 C 2 = 66 ways to pick 2 men (from a pool of 12)
10 C 1 = 10 ways to pick 1 woman (from a pool of 10)

So 66*10 = 660 ways to pick 2 men & 1 woman

note: I'm using the Combination Formula when I refer to something like 12 C 2 (basic template: n C r)

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If we have 3 men, then...

12 C 3 = 220 ways to pick 2 men (from a pool of 12)
10 C 0 = 1 way to pick 0 woman (from a pool of 10)

So 220*1 = 220 ways to pick 3 men & 0 women

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Add up the subtotals: 660+220 = 880

Final Answer: There are 880 ways to pick a committee where the majority must be men.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A group of 10 women and 12 men must select a 3-person committee. How many committees are possible if there must be a majority of men?
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# of ways to succeed:: 12C3 + 12C2*10C1 = 220+142 = 362
# of random sets of 3: 22C3 = 1540
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Ans: 362/1540 = 0.2351
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Cheers,
Stan H.