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# Another combinatorics

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Manager
Joined: 11 Jul 2004
Posts: 120
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Kudos [?]: 2 [0], given: 0

Another combinatorics [#permalink]  17 Aug 2004, 09:27
Can someone please explain a general appraoch to these problems?

5 married couples. Need to select a group of 3 such that no husband and wife are in the group

thanks!
Senior Manager
Joined: 19 May 2004
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Kudos [?]: 8 [0], given: 0

[#permalink]  17 Aug 2004, 09:38
I hope i got it right:

Is it 80 ?

I tried 2 ways:

1) Total - unwanted = 3C10 - 5*8 = 80

2) Count the wanted combinations: 3C5+3C5 + (2C5)*3 + (2C5)*3 = 80
Which is : 3 Women OR 2 Men OR 2 Men and 1 woman OR 2 Women and 1 Man.
Manager
Joined: 11 Jul 2004
Posts: 120
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[#permalink]  17 Aug 2004, 09:44
dookie your answer is CORRECT

can you please explain why unwanted is 5*8?

thanks!
Senior Manager
Joined: 19 May 2004
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[#permalink]  17 Aug 2004, 09:53
For the unwanted, i first chose 1 married couple. There are 5 ways to choose this couple.
Then i chose another person to make it a threesome. There are 8 people left to choose from, since 2 people out of the 10 were already selected.

Now just multiply 5*8.
Senior Manager
Joined: 25 Jul 2004
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[#permalink]  17 Aug 2004, 10:51
Alternatively, you can think of a couple as a single unit (since they can't be chosen together), and you get 5 Choose 3

Now for each item, you choose husband or wife, so you multiply result by 2 for each choice

5C3 * 2^3 = 10 * 8 = 80
[#permalink] 17 Aug 2004, 10:51
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# Another combinatorics

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