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
If there are 4 pairs of twins, and a committee will be
This topic has expert replies
-
- Moderator
- Posts: 7187
- Joined: Thu Sep 07, 2017 4:43 pm
- Followed by:23 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
Here's one approach.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
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
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- 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
GMAT/MBA Expert
- Scott@TargetTestPrep
- GMAT Instructor
- Posts: 7294
- Joined: Sat Apr 25, 2015 10:56 am
- Location: Los Angeles, CA
- Thanked: 43 times
- Followed by:29 members
The number of ways to select the 3 pairs from 4 pairs is 4C3 = 4.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
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
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews