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2 Combination Qs. 1. How many ways to choose a group of 3

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Priti
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2 Combination Qs. 1. How many ways to choose a group of 3 [#permalink] New post   18 Aug 2005, 02:43
2 Combination Qs.

1. How many ways to choose a group of 3 people from 4 couples so that no couples are chosen?

2. A box contains 5 diff. red and 6 diff. white balls. In how many ways can 6 balls be selected so that there are atleast 2 balls of each color?

Please explain steps to get to the answer.

Thanks.



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Re: 2 Combination Qs. 1. How many ways to choose a group of 3 [#permalink] New post   18 Aug 2005, 03:53
Priti wrote:
2 Combination Qs.

1. How many ways to choose a group of 3 people from 4 couples so that no couples are chosen?


Select 3 couples and then choose 1 from each.

C(4,3) * C(2,1) * C(2,1) * C(2,1) = 4 * 2 * 2 * 2= 32
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Re: 2 Combination Qs. 1. How many ways to choose a group of 3 [#permalink] New post   18 Aug 2005, 03:58
Quote:
2. A box contains 5 diff. red and 6 diff. white balls. In how many ways can 6 balls be selected so that there are atleast 2 balls of each color?


We need to sum up mutually exclusive combos.

2 of one kind, 4 of another ---- A
3 each --- B

A = C(5,2) * C (6,4) + C (5,4) + C (6,2) = 10 * 15 + 5 * 15 = 225
B = C (5,3) * C (6,3) = 10 * 20 = 200

Total = 425
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Re: 2 Combination Qs. 1. How many ways to choose a group of 3 [#permalink] New post   18 Aug 2005, 05:57
1) I couldn't come up with a quick solution for the first problem without looking at ajb's post. I agree with him on the approach.

2) I did basically the same thing as ajb, only the other way around. I considered

(All possibilities) - (0 red, six white) - (1 red, 5 white) - (1 white, 5 red) =
C(6,11) - C(0,5)*C(6,6) - C(1,5)*C(5,6) - C(1,6)*C(5,5) = 425
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Re: 2 Combination Qs. 1. How many ways to choose a group of 3 [#permalink] New post   18 Aug 2005, 06:17
Priti wrote:
2 Combination Qs.

1. How many ways to choose a group of 3 people from 4 couples so that no couples are chosen?


Total combinations of choosing 3 out of 8 (order not important): 8!/(5!*3!) = 56
Combinations for groups with couples: 4 couples * 6 (one from each other couple) = 24
Total combinations - Combinations for groups with couples = 32
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Re: 2 Combination Qs. 1. How many ways to choose a group of 3 [#permalink] New post   18 Aug 2005, 06:27
1. Three people can be chosen from a group of 8(4 couples) in 8C3 ways = 56 ways
Number of ways to choose 1 couple from a group of 4 is 4 * 6 - after choosing 1 couple there are 6 different ways to choose the other person.
So number of ways to choose a group of 3 people so that no couples are chosen will be 56 - 24 = 32 ways.

2.
2R + 4W = 5C2 * 6C4 = 10 * 15 = 150
4R + 2W = 5C4 * 6C2 = 5 * 15 = 75
3R + 3W = 5C3 * 6C3 = 10 * 20 = 200
which is 425 ways
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Re: 2 Combination Qs. 1. How many ways to choose a group of 3 [#permalink] New post   18 Aug 2005, 09:52
rthothad wrote:
1. Three people can be chosen from a group of 8(4 couples) in 8C3 ways = 56 ways
Number of ways to choose 1 couple from a group of 4 is 4 * 6 - after choosing 1 couple there are 6 different ways to choose the other person.
So number of ways to choose a group of 3 people so that no couples are chosen will be 56 - 24 = 32 ways.

2.
2R + 4W = 5C2 * 6C4 = 10 * 15 = 150
4R + 2W = 5C4 * 6C2 = 5 * 15 = 75
3R + 3W = 5C3 * 6C3 = 10 * 20 = 200
which is 425 ways


plz rthothad i don 't understand why you do retrieve 24 from 56; Plz can you explain why number of ways to choose a group of 3 people so that no couples are chosen will be 56 - 24 thanks

regards


mandy
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ranga41
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Re: 2 Combination Qs. 1. How many ways to choose a group of 3 [#permalink] New post   18 Aug 2005, 10:28
Q1)

8*6*4/3! = 32 (8 choices for first, 6 for second, 4 for 3rd and divide by 6 to get rid of duplicates, since order does not matter)

Q2) 425, Good explanations above
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rthothad
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Re: 2 Combination Qs. 1. How many ways to choose a group of 3 [#permalink] New post   18 Aug 2005, 19:44
mandy wrote:
plz rthothad i don 't understand why you do retrieve 24 from 56; Plz can you explain why number of ways to choose a group of 3 people so that no couples are chosen will be 56 - 24 thanks

regards


mandy


mandy, My initial 8C3 combination contains couples also, so in order to remove all the combination that contains a couple, I had to deduct. Hope I am clear.



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