Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 563 reviewsTarget Test Prep

combinations problem

Problem Solving — algebra and arithmetic
Expert replies
Post new topic Post Reply
Subscribe topic
Previous Topic Next Topic
Newbie | Next Rank: 10 Posts
Posts: 6
Joined: Wed Jun 23, 2010 10:46 am
by ponnu_2k » Mon Jul 12, 2010 2:56 pm
Can you please explain the following:

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

Thanks,
PJ
Top
Join the discussion
Source: — Quantitative Reasoning |
User avatar
Master | Next Rank: 500 Posts
Posts: 324
Joined: Mon Jul 05, 2010 6:44 amFollowed by: 3 members
Location: London
Thanked: 70 times

by kmittal82 » Tue Jul 13, 2010 2:27 am
Good question.

Let the couples be a1a2, b1b2, c1c2, d1d1 and e1e2 i.e. 10 people

Now, we can chose 1 out of those in 10 ways. Lets say we chose a1
For the next person, we have 9 to chose from, but we cant chose a2, hence we only really have 8 we can chose from. Lets say we picked b1
For the last place, we have 8 people remaining, out of which 2 can't be chosen to fulfill the condition, so we have 6 to chose from (we already have a1 and b1, we can't pick a2 or b2, hence 2 out of the remaining options are invalid)


Total ways of chosing = 10 x 8 x 6 = 480
Each of those 3 people can be further rearranged in 6 ways

Therefore, number of committees = 480/6 = 80

(D) should be the answer
Top
Join the discussion
Senior | Next Rank: 100 Posts
Posts: 36
Joined: Thu Dec 31, 2009 5:54 am
Location: New York City
Thanked: 2 times

by bdevas01 » Wed Jul 14, 2010 5:01 am
Note: This is a combinatorics problem; you should know what cominations and permutations are in order to solve this problem correctly.

Now, think of it like this:

Number of committees without a married couple = total number of committees possible - number of committees with a married couple.

To compute the total number of committees possible, we use the combinations formula: 10C3 = ( (10!) / (3! * 7!) ), which equals 120.

Now we need to know the number of committees, out of the 120 possibilities, that contain married couples. We know that if a committee is going to have a married couple, it has to be one of the 5 couples plus one of the 8 remaining individuals. Therefore, the number of committees with a married couple is 5 x 8, which equals 40.

Now we can plug these values into our original equation:

Number of committees without a married couple = 120 - 40.

The answer is (D) 80.

Let me know if any of this is unclear to you.
Top
Join the discussion
Master | Next Rank: 500 Posts
Posts: 385
Joined: Sun Jul 12, 2009 10:16 pmFollowed by: 2 members
Thanked: 29 times
GMAT Score:710

by debmalya_dutta » Thu Jul 15, 2010 10:13 am
kmittal82 wrote:Good question.

Let the couples be a1a2, b1b2, c1c2, d1d1 and e1e2 i.e. 10 people

Now, we can chose 1 out of those in 10 ways. Lets say we chose a1
For the next person, we have 9 to chose from, but we cant chose a2, hence we only really have 8 we can chose from. Lets say we picked b1
For the last place, we have 8 people remaining, out of which 2 can't be chosen to fulfill the condition, so we have 6 to chose from (we already have a1 and b1, we can't pick a2 or b2, hence 2 out of the remaining options are invalid)


Total ways of chosing = 10 x 8 x 6 = 480
Each of those 3 people can be further rearranged in 6 ways

Therefore, number of committees = 480/6 = 80

(D) should be the answer
Hi Guys,
My understanding was that the total ways of choosing = 10 x 8 x 6 = 480 does not include the 6 ways in which the 3 people can be arranged . Is that the case though ?
Top
Join the discussion
Senior | Next Rank: 100 Posts
Posts: 36
Joined: Thu Dec 31, 2009 5:54 am
Location: New York City
Thanked: 2 times

by bdevas01 » Thu Jul 15, 2010 10:33 am
debmalya_dutta wrote:
kmittal82 wrote: Hi Guys,
My understanding was that the total ways of choosing = 10 x 8 x 6 = 480 does not include the 6 ways in which the 3 people can be arranged . Is that the case though ?
Debmalya,

Actually, the 480 DOES include the 6 ways that the 3 people can be rearranged. That's actually why kmittal divided the number by 4.
Last edited by bdevas01 on Thu Jul 15, 2010 11:40 am, edited 1 time in total.
Top
Join the discussion
Master | Next Rank: 500 Posts
Posts: 385
Joined: Sun Jul 12, 2009 10:16 pmFollowed by: 2 members
Thanked: 29 times
GMAT Score:710

by debmalya_dutta » Thu Jul 15, 2010 11:28 am
So bdevas01 ,
480 is the answer ? If the arrangement in the group would be applicable (hypothetical scenario) sense then the total number of ways would have been 480 * 6....right ?
Top
Join the discussion
Senior | Next Rank: 100 Posts
Posts: 36
Joined: Thu Dec 31, 2009 5:54 am
Location: New York City
Thanked: 2 times

by bdevas01 » Thu Jul 15, 2010 11:42 am
debmalya_dutta wrote:So bdevas01 ,
480 is the answer ? If the arrangement in the group would be applicable (hypothetical scenario) sense then the total number of ways would have been 480 * 6....right ?
That was my mistake, I thought u were talking about something else. I edited my previous post to reflect the correct information.

You are correct in saying 480 includes the 6 ways to arrange each person so there is no need to multiply 480 * 6. Any further questions?
Top
Join the discussion
Master | Next Rank: 500 Posts
Posts: 379
Joined: Wed Jun 03, 2009 3:05 amFollowed by: 1 member
Thanked: 19 times
GMAT Score:690

by sreak1089 » Fri Jul 16, 2010 12:31 am
5 couples => 10 people.

3-member committee needs to be formed, which can be done in 10C3 ways.
However, it includes married couples. 1 married couple can be selected in 5C1 ways. Another
one person can be selected in 8C1 ways.

thus, the answer is 10C3 - 5C1 * 8C1 = 120 - 5*8 = 80.
hence d
Top
Join the discussion