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# Q. A group of 3 people is to be selected from 5 married

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Q. A group of 3 people is to be selected from 5 married [#permalink]  08 May 2008, 09:34
Q. A group of 3 people is to be selected from 5 married couples, such that the group does not include two people who are married to each other. How many such groups are possible?

A. One approach which i understand is-

3 people can be selected from 10 in 10c3 ways= 120
A group of 3 with one married couple can be selected in= 5c1 * 8c1= 40 ways.

Hence the total number of ways to select a group of 3 such that no 2 people selected are married to each other= 120-40=80

My question is-

why doesnt 10c1 * 8c1 * 6c1 work as an answer??
Intern
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Re: Permutation & Combination [#permalink]  18 May 2008, 09:31
where are the big brains of this forum?
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Re: Permutation & Combination [#permalink]  18 May 2008, 09:39
Let me give it a try. You cannot select 3 people from the same group in 10C1 * 8C1 * 6C1 number of ways.

If this approach were correct, then any 3 people can be selected in
10C1*9C1*8C1 ways, which is clearly incorrect as you mentioned that the correct ans for this is 10C3

10 * 8 * 6 will essentially give u the number of ways of selecting 1 person from DIFFERENT groups of 10, 8 and 6 people.

Hope this helps.
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Re: Permutation & Combination [#permalink]  18 May 2008, 10:07
Expert's post
1. We can choose any person of 5*2 people: $$C^{10}_1$$

2. We can choose any person of 5*2-1 people but one that is a spouse for the first-chosen person: $$C^{8}_1$$

3. We can choose any person of 5*2-2 people but two that are spouses for the first-chosen and second-chosen person: $$C^{6}_1$$

4. Now, we have exclude order in our group: N=$$\frac{C^{10}_1*C^{8}_1*C^{6}_1}{P^3_3}=80$$
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Re: Permutation & Combination   [#permalink] 18 May 2008, 10:07
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