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by GMATGuruNY » Mon Dec 03, 2018 5:09 am
AAPL wrote:GMAT Prep

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
Number of options for the 1st person = 8.
Number of options for the 2nd person = 6. (Of the 7 people left, we can't use the twin of the 1st person chosen, leaving 7-1= 6 choices.)
Number of options for the 3rd person = 4. (Of the 6 people left, we can't use the twins of the 2 people already chosen, leaving 6-2 = 4 choices.)

Since the ORDER of the selections doesn't matter -- ABC is the same COMMITTEE as BCA -- we divide by the number of ways to arrange the 3 people chosen (3!):
(8*6*4)/3! = 32.

The correct answer is A.
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by Brent@GMATPrepNow » Mon Dec 03, 2018 7:24 am
AAPL wrote:GMAT Prep

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.
Here's another 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, we 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

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

by fskilnik@GMATH » Mon Dec 03, 2018 12:03 pm
AAPL wrote:GMAT Prep

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
$$\left. \matrix{
\# \,\,{\rm{total}}\,\,{\rm{committees}}\,\,{\rm{ = }}\,\,\,{\rm{C}}\left( {8,3} \right) = {{8 \cdot 7 \cdot 6} \over {3!}} = 56 \hfill \cr
\# \,\,{\rm{committees}}\,\,{\rm{with}}\,\,{\rm{siblings}}\,\,{\rm{ = }}\,\,\,\underbrace {\,\,\,4\,\,\,}_{{\rm{choice}}\,\,{\rm{of}}\,\,{\rm{pair}}\,\,{\rm{of}}\,\,{\rm{twins}}} \cdot \underbrace {\,\,\,6\,\,}_{{\rm{choice}}\,\,{\rm{out - of - the - pair}}} = 24\,\,\, \hfill \cr} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,?\,\,\, = \,\,\,56 - 24 = 32$$


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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by Scott@TargetTestPrep » Thu Apr 04, 2019 5:16 pm
AAPL wrote:GMAT Prep

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

Since there can be no siblings on the committee, 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, so there are (8 x 6 x 4)/3! = 8 x 4 = 32 possible committees.

Answer: A

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