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macattack: Senior | Next Rank: 100 Posts; Posts: 90; Joined: Wed Jun 26, 2013 10:55 pmFollowed by: 1 member; Thanked: 11 times

by macattack » Tue Aug 13, 2013 1:24 am

Help please

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Source: Beat The GMAT — Problem Solving | Tue Aug 13, 2013 1:24 am

macattack: Senior | Next Rank: 100 Posts; Posts: 90; Joined: Wed Jun 26, 2013 10:55 pmFollowed by: 1 member; Thanked: 11 times

by macattack » Tue Aug 13, 2013 1:26 am

TO view the image properly click on open image in new tab... thx

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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 » Tue Aug 13, 2013 2:19 am

A committee of three people is to be chosen from four married couples. What is the number of different commities that can be chosen if two people who are married to each other cannot both serve on this committee?

16
24
26
30
32

Number of options for the 1st person = 8.
Number of options for the 2nd person = 6. (Of the 7 remaining people, anyone but the spouse of the first person selected.)
Number of options for the 3rd person = 4. (Of the 6 remaining people, anyone but the spouses of the first 2 people selected.)
To combine these options, we multiply:
8*6*4.

Since the ORDER of the selections doesn't matter -- ABC is the same committee as BCA -- the product above must be divided by the number of ways to ARRANGE the 3 people chosen (3!):
(8*6*4)/(3*2*1) = 32.

The correct answer is E.

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macattack: Senior | Next Rank: 100 Posts; Posts: 90; Joined: Wed Jun 26, 2013 10:55 pmFollowed by: 1 member; Thanked: 11 times

by macattack » Tue Aug 13, 2013 2:23 am

Thanks that was really helpful. However i would like to know how you got the 3!. What is the mathematical rule behind it. THanks.

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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 » Tue Aug 13, 2013 2:39 am

macattack wrote:Thanks that was really helpful. However i would like to know how you got the 3!. What is the mathematical rule behind it. THanks.

Check my SECOND post here:

https://www.beatthegmat.com/counting-problem-t72690.html

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Tue Aug 13, 2013 7:11 am

A committee of three people is to be chosen from four married couples. What is the number of different committee 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

Here's another 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)

If anyone is interested, we have a free video on calculating combinations (like 4C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

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: https://www.gmatprepnow.com/module/gmat-counting?id=775

Brent Hanneson - Creator of GMATPrepNow.com

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