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

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

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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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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: http://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: http://www.gmatprepnow.com/module/gmat-counting/video/775

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

Then you can try solving the following questions:

EASY
- http://www.beatthegmat.com/what-should-be-the-answer-t267256.html
- http://www.beatthegmat.com/counting-problem-company-recruitment-t244302.html
- http://www.beatthegmat.com/picking-a-5-digit-code-with-an-odd-middle-digit-t273110.html
- http://www.beatthegmat.com/permutation-combination-simple-one-t257412.html
- http://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- http://www.beatthegmat.com/combinatorics-solution-explanation-t273194.html
- http://www.beatthegmat.com/arabian-horses-good-one-t150703.html
- http://www.beatthegmat.com/sub-sets-probability-t273337.html
- http://www.beatthegmat.com/combinatorics-problem-t273180.html
- http://www.beatthegmat.com/digits-numbers-t270127.html
- http://www.beatthegmat.com/doubt-on-separator-method-t271047.html
- http://www.beatthegmat.com/combinatorics-problem-t267079.html


DIFFICULT
- http://www.beatthegmat.com/wonderful-p-c-ques-t271001.html
- http://www.beatthegmat.com/permutation-and-combination-t273915.html
- http://www.beatthegmat.com/permutation-t122873.html
- http://www.beatthegmat.com/no-two-ladies-sit-together-t275661.html
- http://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent

_________________
Brent Hanneson – Creator of GMATPrepNow.com
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Post
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

_________________

Scott Woodbury-Stewart
Founder and CEO
scott@targettestprep.com



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