* Combinatorics --> committee containing married people

[ricms08]

If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

A. 20
B. 40
C. 50
D. 80
E. 120

The answer to this is D.

I figure it this way:

10C3=120

5*1*8=40

120-40=80

IS taking 5*1*8 right for the total number of combinations with two married people?

Thanks

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[reva.maheshwari]

Please explain how is 5 * 8 * 1 is considered for calculation.

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[Ben Ku]

Please cite the source (author) of this problem. We cannot reply unless a source is cited (and, if no source is cited, we will have to delete the post!). Thanks.

[SeanM921]

It source is the GMATPrep problem QPS00113.

Please cite the source (author) of this problem. We cannot reply unless a source is cited (and, if no source is cited, we will have to delete the post!). Thanks.

[SeanM921]

I solved this problem by noticing that there are 10 individuals that can be chosen (5 married people)

m m m m m
w w w w w

I wrote out the number of men and women (see above) and counted the number of people one individual can be on the board with. Each individual can be on the board with 8 people. (He/she cannot be on the board with his/her spouse or himself/herself)

Then I multiplied the number of possible people a person can be on the board with (8), with the total number of people (10).

8 * 10 = 80

Will using this method always get me to the correct answer?

[RonPurewal]

nope. blind coincidence. (if there are 8 people on the committee, then there are 32 possible committees, but this approach would say 48.)

for starters, this method makes NO use of the fact that there are three people on a committee. just doesn't use that fact at all.
that's clearly an essential feature; ergo, this method doesn't work.

[RonPurewal]

more constructively-- the actual problems with that method are twofold:

1/
(as mentioned above)
it doesn't account for the third committee member.

2/
it doesn't account for repeated possibilities.
e.g., if A and B are individuals who aren't married to each other, then that method counts "A, B" and "B, A" as two different pairs of people rather than one.

taking these facts together, we can figure out what your method actually DOES do. specifically, your method gives the possible numbers of committees of two people (excluding married couples), if order matters (e.g., if one of the two people is the chairperson/senior member).

[SeanM921]

Okay, Thanks.

[RonPurewal]

you're welcome.

[gayatri.k]

I am able to get to 120, as the total number of possible combinations without the rule that no one individual can be selected with their spouse:

10!/(3!7!)

However, I am not sure where to go from there. How did you get to 80?

Apologies if this is answered above - I read through the responses and it wasn't clear to me. I'd really appreciate if someone could explain this again.

Thank you!

[RonPurewal]

if you want to do it that way ^^ then you have to subtract out the number of threesomes that DO include a married couple (i.e., the committees that are excluded by the rule against married couples).

how many of those are there?
well, we have ...
couple #1 plus any of the 8 remaining people
couple #2 plus any of the 8 remaining people
couple #3 plus any of the 8 remaining people
couple #4 plus any of the 8 remaining people
couple #5 plus any of the 8 remaining people
...so, 40 excluded possibilities. 120 – 40 = 80 legitimate possibilities.

[gayatri.k]

Thank you, Ron!

[RonPurewal]

sure.