BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Combinations problem.

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 391
Joined: Sat Mar 02, 2013 5:13 am
Thanked: 50 times
Followed by:4 members

Combinations problem.

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?

3
4
5
6
8
Top
Source: — Problem Solving |
User avatar
Master | Next Rank: 500 Posts
Posts: 490
Joined: Thu Jul 04, 2013 7:30 am
Location: Chennai, India
Thanked: 83 times
Followed by:5 members

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?

3
4
5
6
8

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.
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
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?

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.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top

GMAT/MBA Expert

User avatar
Elite Legendary Member
Posts: 10392
Joined: Sun Jun 23, 2013 6:38 pm
Location: Palo Alto, CA
Thanked: 2867 times
Followed by:511 members
GMAT Score:800

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.

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
Image
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 490
Joined: Thu Jul 04, 2013 7:30 am
Location: Chennai, India
Thanked: 83 times
Followed by:5 members

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.
Top
User avatar
Legendary Member
Posts: 1556
Joined: Tue Aug 14, 2012 11:18 pm
Thanked: 448 times
Followed by:34 members
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
Top
Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Tue Oct 15, 2013 2:59 am
Thanked: 1 times
Followed by:1 members

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)
Top
Post new topic Post Reply
• Page 1 of 1