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There are four distinct pairs of brothers and sisters. In

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There are four distinct pairs of brothers and sisters. In [#permalink] New post 27 Jan 2008, 10:25
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There are four distinct pairs of brothers and sisters. In how many ways can a committee of 3 be formed and NOT have siblings in it?
 8
 24
 32
 56
 80


How can we solve this one directly with combs?
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Re: combinations and permutations [#permalink] New post 28 Jan 2008, 03:59
Consider each pair a distinct object.
4C3 = 4
Then multiply by 2 for each person chosen since each object has 2 people to choose from.
2*2*2 = 8
4*8 = 32

Answer C
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Re: combinations and permutations [#permalink] New post 28 Jan 2008, 06:17
eschn3am wrote:
Consider each pair a distinct object.
4C3 = 4
Then multiply by 2 for each person chosen since each object has 2 people to choose from.
2*2*2 = 8
4*8 = 32

Answer C


Good solution, eschn3am. Since marcodonzelli wants a combinations formula, here it is: 4C3*(2C1)^3
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Re: combinations and permutations [#permalink] New post 28 Jan 2008, 09:55
See this link for more info:

7-p395941
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Re: combinations and permutations [#permalink] New post 28 Jan 2008, 10:15
dzyubam wrote:
See this link for more info:

7-p395941


what do you all think about this approach:

8c3 are the total number of combinations
4c1 represent the choice of 1 couple od siblings
6 are the people who can be chosen to occupy the third place

8c3-4c1*6=56-24=32

moreover do you think about this other way?

Pick one person out of 8, then one out of 6 (the first person with her or his sibling excluded), then one out of 4, giving 8*6*4 permutations. These are called permutations because we count every distinct combination of 3 people 6 times (6=3!). For example, we count {Ann, Bill, and Jane} as {Ann, Bill, Jane}, {Ann, Jane, Bill}, {Jane, Ann, Bill}, {Jane, Bill, Ann}, {Bill, Ann, Jane}, {Bill, Jane, Ann}. To get the number of combinations, we divide the permutations by 6, to get (8*6*4)/6 = 32 combinations

I am not sure about its quickness.
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Re: combinations and permutations [#permalink] New post 31 Oct 2009, 18:24
My approach is to choose the total possible number of ways and exclude undesired ones.

The number of ways to choose 3 people out of 8 is C(8,3), order does not

and the number of ways to form a group of 3 from a married couple + 6 more people, it can be restated as in how many ways we can select one person out of 6 remaining ones. The total number of ways to select 1 person out of 6 and there are 4 different married couples.

Soln: C(8,3) - 4*C(6,1) =32
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Re: combinations and permutations   [#permalink] 31 Oct 2009, 18:24
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