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#1
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Simple Question
For some reason I can't get the right answer for this.
Suppose a vase has 15 balls, of which 8 are red and seven are black. In how many ways can 5 balls be chosen so that at least two are red? |
#2
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Re: Simple Question
[ QUOTE ]
For some reason I can't get the right answer for this. Suppose a vase has 15 balls, of which 8 are red and seven are black. In how many ways can 5 balls be chosen so that at least two are red? [/ QUOTE ] C(8,2)*C(7,3)+C(8,3)*C(7,2)+C(8,4)*C(7,1)+C(8,5) This assumes it doesn't matter which order the balls are chosen. |
#3
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Re: Simple Question
This question is insufficiently clear.
Do you mean: 8 chose 2 * 7 chose 3 + 8 chose 3 * 7 chose 2 + 8 chose 4 * 7 chose 1 + 8 chose 5 * 7 chose 0 = 2702 ? |
#4
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Re: Simple Question
[ QUOTE ]
For some reason I can't get the right answer for this. Suppose a vase has 15 balls, of which 8 are red and seven are black. In how many ways can 5 balls be chosen so that at least two are red? [/ QUOTE ] C(15,5) - C(7,5) - 8*C(7,4) = 2702. |
#5
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Re: Simple Question
Thanks alot, I forgot to subtract out all combinations of 5 black.
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