Welcome! Check out our free B-School Guides to learn how you compare with other applicants.
Login or Register
 

GMAT PREP PS question

This topic has 2 expert replies and 1 member reply
Bookmark
Previous Topic Next Topic
alex.gellatly GMAT Destroyer! Default Avatar
Joined
16 Nov 2011
Posted:
435 messages
Followed by:
16 members
Thanked:
46 times
GMAT PREP PS question Post Mon Apr 16, 2012 1:17 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    A committee of three 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?

    (A) 16
    (B) 24
    (C) 26
    (D) 30
    (E) 32

    Thanks in advanced

    Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums!
    killer1387 GMAT Destroyer! Default Avatar
    Joined
    03 Apr 2011
    Posted:
    583 messages
    Followed by:
    4 members
    Thanked:
    52 times
    Post Mon Apr 16, 2012 1:24 am
    alex.gellatly wrote:
    A committee of three 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?

    (A) 16
    (B) 24
    (C) 26
    (D) 30
    (E) 32

    Thanks in advanced
    total no of ways in which committee can be formed = 8c3=56
    no. of ways in which one couple is there and any one among remaining 6 is present = 4c1*6c1=24

    required probability = 56-24 = 32

    Hence E

    Thanked by: jsockol

    GMAT/MBA Expert

    Anurag@Gurome GMAT Instructor
    Joined
    02 Apr 2010
    Posted:
    3835 messages
    Followed by:
    466 members
    Thanked:
    1755 times
    GMAT Score:
    770
    Post Mon Apr 16, 2012 3:45 am
    alex.gellatly wrote:
    A committee of three 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?

    (A) 16
    (B) 24
    (C) 26
    (D) 30
    (E) 32

    Thanks in advanced
    Number of ways to choose 3 couples from 4 couples = 4C3 = 4

    Now each of these 3 couples can send two persons (husband or wife), number of ways of doing this = 2 * 2 * 2 = 2^3 = 8

    Therefore, total number of ways: 4C3 * 2^3 = 32

    The correct answer is E.

    _________________
    Anurag Mairal, Ph.D., MBA
    GMAT Expert, Admissions and Career Guidance
    Gurome, Inc.
    1-800-566-4043 (USA)

    Join Our Facebook Groups
    GMAT with Gurome
    https://www.facebook.com/groups/272466352793633/
    Admissions with Gurome
    https://www.facebook.com/groups/461459690536574/
    Career Advising with Gurome
    https://www.facebook.com/groups/360435787349781/

    Thanked by: alex.gellatly

    GMAT/MBA Expert

    GMATGuruNY GMAT Instructor
    Joined
    25 May 2010
    Posted:
    6232 messages
    Followed by:
    1068 members
    Thanked:
    5440 times
    GMAT Score:
    790
    Most Active Expert Most Responsive Expert Most Thanked Expert
    Post Mon Apr 16, 2012 3:57 am
    alex.gellatly wrote:
    A committee of three 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?

    (A) 16
    (B) 24
    (C) 26
    (D) 30
    (E) 32

    Thanks in advanced
    An alternate approach:

    Number of choices for the 1st person = 8.
    Number of choices for the 2nd person = 6. (7 people left, but we can't choose the spouse of the first committee member chosen, leaving 7-1= 6 choices.)
    Number of choices for the 3rd person = 4. (6 people left, but we can't choose the spouses of the first 2 committee members chosen, leaving 6-2 = 4 choices.)

    To combine the number of choices we have for each position on the committee, we multiply the numbers above:
    8*6*4.

    But the order in which the committee members are chosen doesn't matter: selecting ABC will yield the same committee as selecting BAC.
    So that we don't overcount these duplicate combinations, we must divide by the number of ways to arrange the 3 members chosen (3!).

    Thus:
    Total number of possible committees = (8*6*4)/3! = 32.

    The correct answer is E.

    _________________
    Mitch Hunt
    GMAT Private Tutor and Instructor
    GMATGuruNY@gmail.com
    If you find one of my posts helpful, please take a moment to click on the "Thank" icon.
    Contact me about long distance tutoring!

    Thanked by: alex.gellatly
    Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now.

    Best Conversation Starters
    1 vipulgoyal 32 topics
    2 hemant_rajput 23 topics
    3 guerrero 20 topics
    4 abhirup1711 17 topics
    5 sana.noor 15 topics
    See More Top Beat The GMAT Members...

    Most Active Experts
    1 image description Brent@GMATPrepNow

    GMAT Prep Now Teacher

    		 204 posts
    2 image description GMATGuruNY

    The Princeton Review Teacher

    		 144 posts
    3 image description Jim@StratusPrep

    Stratus Prep

    		 54 posts
    4 image description Jon@Admissionado

    Admissionado

    		 42 posts
    5 image description David@VeritasPrep

    Veritas Prep

    		 35 posts
    See More Top Beat The GMAT Experts