If there are 4 pairs of twins, and a committee will be

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

by BTGmoderatorDC » Tue Mar 19, 2019 9:04 pm

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If there are 4 pairs of twins, and a committee will be formed with 3 members. In how many ways this committee formed in a way that no siblings in a group?

A. 32
B. 24
C. 56
D. 44
E. 40

OA A

Source: GMAT Prep

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Source: Beat The GMAT — Problem Solving | Tue Mar 19, 2019 9:04 pm

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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 » Wed Mar 20, 2019 5:04 am

BTGmoderatorDC wrote:If there are 4 pairs of twins, and a committee will be formed with 3 members. In how many ways this committee formed in a way that no siblings in a group?

A. 32
B. 24
C. 56
D. 44
E. 40

OA A

Source: GMAT Prep

Here's one approach.

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

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

If anyone is interested, I have a 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 pairs of twins and choose 1 person to be on the committee.
There are 2 people, so this stage can be accomplished in 2 ways.

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

Stage 4: Take one of the 3 selected pairs of twins and choose 1 person to be on the committee.
There are 2 people, 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: A

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat- ... /video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html

MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
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- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html

DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8151; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Fri Mar 22, 2019 8:12 am

BTGmoderatorDC wrote:If there are 4 pairs of twins, and a committee will be formed with 3 members. In how many ways this committee formed in a way that no siblings in a group?

A. 32
B. 24
C. 56
D. 44
E. 40

OA A

Source: GMAT Prep

The number of ways to select the 3 pairs from 4 pairs is 4C3 = 4.

Since there can be no siblings on the board each twin can be selected in 2C1 ways, so:

2C1 x 2C1 x 2C1 = 2 x 2 x 2 = 8

So the total number of ways to select the committee is 4 x 8= 32.

Alternate Solution:

For the first member, there are 8 choices. Since the sibling of the first member cannot be chosen, there are 6 choices for the second member. By the same logic, there are 4 choices for the last member. Notice that the 8 x 6 x 4 choices count each committee 3! times; therefore, there are (8 x 6 x 4)/3! = 8 x 4 = 32 possible committees.

Answer: A

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