• EMPOWERgmat Slider
    1 Hour Free
    BEAT THE GMAT EXCLUSIVE

    Available with Beat the GMAT members only code

    MORE DETAILS
    EMPOWERgmat Slider
  • Target Test Prep
    5-Day Free Trial
    5-day free, full-access trial TTP Quant

    Available with Beat the GMAT members only code

    MORE DETAILS
    Target Test Prep
  • Magoosh
    Magoosh
    Study with Magoosh GMAT prep

    Available with Beat the GMAT members only code

    MORE DETAILS
    Magoosh
  • Kaplan Test Prep
    Free Practice Test & Review
    How would you score if you took the GMAT

    Available with Beat the GMAT members only code

    MORE DETAILS
    Kaplan Test Prep
  • PrepScholar GMAT
    5 Day FREE Trial
    Study Smarter, Not Harder

    Available with Beat the GMAT members only code

    MORE DETAILS
    PrepScholar GMAT
  • Varsity Tutors
    Award-winning private GMAT tutoring
    Register now and save up to $200

    Available with Beat the GMAT members only code

    MORE DETAILS
    Varsity Tutors
  • Veritas Prep
    Free Veritas GMAT Class
    Experience Lesson 1 Live Free

    Available with Beat the GMAT members only code

    MORE DETAILS
    Veritas Prep
  • e-gmat Exclusive Offer
    Get 300+ Practice Questions
    25 Video lessons and 6 Webinars for FREE

    Available with Beat the GMAT members only code

    MORE DETAILS
    e-gmat Exclusive Offer
  • Economist Test Prep
    Free Trial & Practice Exam
    BEAT THE GMAT EXCLUSIVE

    Available with Beat the GMAT members only code

    MORE DETAILS
    Economist Test Prep

permutation

This topic has 3 expert replies and 3 member replies
Manjareev Junior | Next Rank: 30 Posts Default Avatar
Joined
08 Apr 2012
Posted:
18 messages

permutation

Post Sat Sep 22, 2012 8:33 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    A committee of 3 people is to be chosen from four married couples.
    what is the number of different committees that can be chosen if two people who are
    married to each other cannot both serve on the committee?

    16
    24
    26
    30
    32

    Pls explain the answer.

    Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums!
    Post Sat Sep 22, 2012 9:13 pm
    Manjareev wrote:
    A committee of 3 people is to be chosen from four married couples.
    what is the number of different committees that can be chosen if two people who are
    married to each other cannot both serve on the committee?

    16
    24
    26
    30
    32

    Pls explain the answer.
    Here's one approach.

    Take the task of selecting the 3 committee members and break it into stages.

    Stage 1: Select the 3 couples from which we will select 1 spouse each.
    There are 4 couples, and we must select 3 of them. Since the order in which we select the 3 couples does not matter, this stage can be accomplished in 4C3 ways (4 ways)

    Stage 2: Take one of the 3 selected couples and choose 1 person to be on the committee.
    There are 2 people in the couple, so this stage can be accomplished in 2 ways.

    Stage 3: Take one of the 3 selected couples and choose 1 person to be on the committee.
    There are 2 people in the couple, so this stage can be accomplished in 2 ways.

    Stage 4: Take one of the 3 selected couples and choose 1 person to be on the committee.
    There are 2 people in the couple, so this stage can be accomplished in 2 ways.

    By the Fundamental Counting Principle (FCP) we can complete all 4 stages (and thus create a 3-person committee) in (4)(2)(2)(2) ways (= 32 ways)

    Answer = E

    Cheers,
    Brent

    Aside: For more information about the FCP, we have a free video on the subject: http://www.gmatprepnow.com/module/gmat-counting?id=775

    _________________
    Brent Hanneson – Founder of GMATPrepNow.com
    Use our video course along with Beat The GMAT's free 60-Day Study Guide

    Check out the online reviews of our course
    Come see all of our free resources

    Thanked by: Manjareev
    GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months!
    Manjareev Junior | Next Rank: 30 Posts Default Avatar
    Joined
    08 Apr 2012
    Posted:
    18 messages
    Post Sat Sep 22, 2012 11:12 pm
    Thank you so much Sir for ur reply. ur explanation is clear. i was just confused for the stage 1 which u had mentioned in the answer. now i know my error of reasoning. thanks.

    LalaB Master | Next Rank: 500 Posts
    Joined
    08 Dec 2010
    Posted:
    425 messages
    Followed by:
    7 members
    Thanked:
    56 times
    GMAT Score:
    690
    Post Mon Oct 01, 2012 8:58 am
    four married couples=4*2= 8 people

    we need to get 6 ppl(excluding 2 married) out of 8
    8!/6!/2!=32

    _________________
    Happy are those who dream dreams and are ready to pay the price to make them come true.(c)

    In order to succeed, your desire for success should be greater than your fear of failure.(c)

    Kaustubhk Junior | Next Rank: 30 Posts Default Avatar
    Joined
    27 Feb 2015
    Posted:
    25 messages
    Thanked:
    1 times
    Post Wed May 24, 2017 8:51 am
    HI brent,

    1st member is selected in 8ways

    2nd member in 6 ways (we can't select the spouse)

    3rd Member in 4 ways

    8*6*4
    The order in which the is not important so we must divide by the factorial of the number of choices to eliminate over-counting:


    (8*6*4)/3! = 8*4 = 32

    Post Wed May 24, 2017 9:14 am
    Kaustubhk wrote:
    HI brent,

    1st member is selected in 8ways

    2nd member in 6 ways (we can't select the spouse)

    3rd Member in 4 ways

    8*6*4
    The order in which the is not important so we must divide by the factorial of the number of choices to eliminate over-counting:


    (8*6*4)/3! = 8*4 = 32
    Perfect!

    Cheers,
    Brent

    _________________
    Brent Hanneson – Founder of GMATPrepNow.com
    Use our video course along with Beat The GMAT's free 60-Day Study Guide

    Check out the online reviews of our course
    Come see all of our free resources

    GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months!

    GMAT/MBA Expert

    Post Thu Jun 01, 2017 4:05 pm
    Manjareev wrote:
    A committee of 3 people is to be chosen from four married couples.
    what is the number of different committees that can be chosen if two people who are
    married to each other cannot both serve on the committee?

    16
    24
    26
    30
    32
    We are given that there are four married couples (or 8 people) and we need to determine the number of ways of choosing 3 people in which no two people are a married couple. This is a special combination problem. Before we tackle this problem, let’s review a combination problem with no restrictions.

    With no restrictions, the number of ways of choosing 3 people from 8 people is 8C3, which is calculated as follows:

    8C3 = 8!/[3!(8-3)!] = 8!/[3!5!] = (8 x 7 x 6)/3! = 56

    8, 7, and 6, in the numerator, represent the number of ways the first, second, and third person can be chosen, respectively. We divide the numerator by 3! because in a combination problem we do not care about the order in which the 3 people are chosen.

    However, in this (special combination) problem, 3 people are chosen in which no married couple can serve together on the committee. The first person could be any one of the 8 people. However, once a person is selected, that person’s spouse cannot also be selected for the committee. This reduces the choices for the second person to 6 possible people (one person has already been selected and that person’s spouse now cannot be selected). Once the second person is chosen for the committee, that person’s spouse cannot be chosen. This reduces the number of people who could be chosen as the third person to 4. Therefore, the number of ways of choosing these 3 people is:

    (8 x 6 x 4)/3! = 32

    Thus, there are 32 ways to choose such a committee.

    Answer: E

    _________________

    Scott Woodbury Stewart Founder & CEO
    GMAT Quant Self-Study Course - 500+ lessons 3000+ practice problems 800+ HD solutions
    5-Day Free Trial 5-DAY FREE, FULL-ACCESS TRIAL TTP QUANT

    Best Conversation Starters

    1 Vincen 152 topics
    2 lheiannie07 61 topics
    3 Roland2rule 49 topics
    4 LUANDATO 44 topics
    5 ardz24 40 topics
    See More Top Beat The GMAT Members...

    Most Active Experts

    1 image description Brent@GMATPrepNow

    GMAT Prep Now Teacher

    140 posts
    2 image description Rich.C@EMPOWERgma...

    EMPOWERgmat

    110 posts
    3 image description EconomistGMATTutor

    The Economist GMAT Tutor

    109 posts
    4 image description GMATGuruNY

    The Princeton Review Teacher

    107 posts
    5 image description DavidG@VeritasPrep

    Veritas Prep

    72 posts
    See More Top Beat The GMAT Experts