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#1
10-10-2005, 01:18 PM
 closer2313 Member Join Date: Dec 2004 Posts: 63
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
10-10-2005, 02:29 PM
 Tom1975 Senior Member Join Date: Jun 2005 Posts: 132
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
10-10-2005, 02:34 PM
 Guest Posts: n/a
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
10-10-2005, 03:21 PM
 BruceZ Senior Member Join Date: Sep 2002 Posts: 1,636
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
10-10-2005, 03:24 PM
 closer2313 Member Join Date: Dec 2004 Posts: 63
Re: Simple Question

Thanks alot, I forgot to subtract out all combinations of 5 black.

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