Combinations problem.

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rakeshd347: Master | Next Rank: 500 Posts; Posts: 391; Joined: Sat Mar 02, 2013 5:13 amFollowed by: 4 members; Thanked: 50 times

by rakeshd347 » Mon Oct 14, 2013 6:03 pm

A committee of three students has to be formed. There are five candidates: Jane, Joan, Paul, Stuart, and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many committees are possible?

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Uva@90: Master | Next Rank: 500 Posts; Posts: 490; Joined: Thu Jul 04, 2013 7:30 amFollowed by: 5 members; Location: Chennai, India; Thanked: 83 times

by Uva@90 » Mon Oct 14, 2013 7:02 pm

rakeshd347 wrote:A committee of three students has to be formed. There are five candidates: Jane, Joan, Paul, Stuart, and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many committees are possible?

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Hi Rakesh,
There are 3 ways we can select a team of three members,

Since answers are of single digits we can write all possibilities.

First, {Jane,Paul,Joan}
second, {Jane, Paul, Jessica}
Third , {Joan,Stuart,Jessica}

Experts correct me if I am wrong.

Regards,
Uva.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Mon Oct 14, 2013 9:18 pm

rakeshd347 wrote:A committee of three students has to be formed. There are five candidates: Jane, Joan, Paul, Stuart, and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many committees are possible?

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Jane requires Paul, but Paul does NOT require Jane.

If Jane is selected, the following committees can be formed:
Jane - Paul - Joan
Jane - Paul - Jessica

If Paul is selected, but Jane is NOT selected, the following committee can be formed:
Paul - Joan - Jessica

If NEITHER Jane nor Paul is selected, the following committee can be formed:
Joan - Stuart - Jessica

Total possible committees = 2+1+1 = 4.

The correct answer is B.

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by [email protected] » Mon Oct 14, 2013 9:53 pm

Hi rakeshd347,

While this can certainly be "called" a combinatorics question, you can see from the explanations that "math" was not anyone's preferred approach. In these types of story problems, when the answers are relatively small (as they are here), the most direct/fastest approach is usually to just "list out" the options.

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Uva@90: Master | Next Rank: 500 Posts; Posts: 490; Joined: Thu Jul 04, 2013 7:30 amFollowed by: 5 members; Location: Chennai, India; Thanked: 83 times

by Uva@90 » Mon Oct 14, 2013 10:11 pm

GMATGuruNY wrote:
rakeshd347 wrote:A committee of three students has to be formed. There are five candidates: Jane, Joan, Paul, Stuart, and Jessica. If Paul and Stuart refuse to be in the committee together and Jane refuses to be in the committee without Paul, how many committees are possible?

3
4
5
6
8
Jane requires Paul, but Paul does NOT require Jane.

If Jane is selected, the following committees can be formed:
Jane - Paul - Joan
Jane - Paul - Jessica

If Paul is selected, but Jane is NOT selected, the following committee can be formed:
Paul - Joan - Jessica

If NEITHER Jane nor Paul is selected, the following committee can be formed:
Joan - Stuart - Jessica

Total possible committees = 2+1+1 = 4.

The correct answer is B.

Mitch,
This is excellent catch to remember for these kind of gmat preparation,
Jane requires Paul, but Paul does NOT require Jane.

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theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Mon Oct 14, 2013 11:49 pm

5 Candidates : Jane, Joan, Paul, Stuart, and Jessica
3 to be selected

So, 5C3 - 2C2*3C1 - 2C2*3C1 = 10 - 3 - 3 = 4

Answer [spoiler]{B} [/spoiler]

R A H U L

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b_joy: Newbie | Next Rank: 10 Posts; Posts: 3; Joined: Tue Oct 15, 2013 2:59 amFollowed by: 1 member; Thanked: 1 times

by b_joy » Tue Oct 15, 2013 3:49 am

Another easy way to do it is by subtracting the unwanted combinations from the total possible combinations.
So the answer would be:
5C3- 2C2*3C1-1C1*3C2

where 5C3: total possible selections
2c2*3C1:combinations where Paul and Stuart are selected (which means only 1 out of rest three can be selected next(3C1))
1C1*3C2: combinations where Jane is selected and Paul is not selected (1C1 for selecting Jane and 3C2 for selecting the rest three people excluding Paul)

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