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A combination of three people is to be chosen from 4 married

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A combination of three people is to be chosen from 4 married [#permalink] New post 19 Oct 2007, 21:54
A combination of three people is to be chosen from 4 married couples, What is the number of different committes that can be chosen if two people who married to each other cannot serve both on the committee?

16
24
26
30
32

Here's what I have so far :

Total combination : 8C3 = 112

We now need to subtract couples married to each other from 112

That's where I' get confused.

What's next
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GMAT the final frontie!!!.

VP
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 [#permalink] New post 19 Oct 2007, 23:00
first of all 8C3 = 8!/(5!*3!) = (8*7*6)/(3*2) = 56

then you need to subtract the different ways a committee containing a married couple will be formed:

4C1 = 4 ----> choosing a married couple
6C1 = 6 ----> choosing the last person in the committee.

total

56-4*6 = 32

the answer is (E)

:)
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 [#permalink] New post 20 Oct 2007, 03:27
Alternative approach
4 wives, 4 husbands
There are 4 possible groups of combination: 3 wives, 2 wives and 1 husband, 1 wives and 2 husband, and all 3 husbands.

Calculate the combination of each case separately.

Case 1: 3 wives
4C3 = 4

Case 2: 2 wives and 1 husband
(4C2)(2C1) = 6 x 2 = 12

Case 3: 1 wives and 2 husband
(4C1)(3C2) = 4 x 3 = 12

Case 4: 3 husbands
4C3 = 4

Total = 4 + 12 + 12 + 4 = 32

(E) is the answer. And of course, I like the original approach better.
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 [#permalink] New post 20 Oct 2007, 08:53
Quote:
4C1 = 4 ----> choosing a married couple
6C1 = 6 ----> choosing the last person in the committee.


I don't get how you come up with this equation.
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 [#permalink] New post 20 Oct 2007, 09:53
KillerS got 4C1 & 6C1 by the following reasoning:
Empty commitee spots -
1 2 3

4C1 =
He views married couples as 1 here and uses the combinations forumla to list all the possible ways to pick 1 couple from 4 available. 1 couple will occupy 2 spots in the committee.

6C1 =
Since we have filled 2 spots with a married couple, we have 1 spot open remaining. Since 1 couple (2 persons) have been selected already, we only have 6 persons remaining to choose from in filling the final spot.

(4C1)*(6C1) = 24, so 56-24 = 32
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 [#permalink] New post 20 Oct 2007, 10:02
yuefei wrote:
KillerS got 4C1 & 6C1 by the following reasoning:
Empty commitee spots -
1 2 3

4C1 =
He views married couples as 1 here and uses the combinations forumla to list all the possible ways to pick 1 couple from 4 available. 1 couple will occupy 2 spots in the committee.

6C1 =
Since we have filled 2 spots with a married couple, we have 1 spot open remaining. Since 1 couple (2 persons) have been selected already, we only have 6 persons remaining to choose from in filling the final spot.

(4C1)*(6C1) = 24, so 56-24 = 32


Perfect ! couldn't write this better myself

Thanks

:)
  [#permalink] 20 Oct 2007, 10:02
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