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Re: How many ways can you group 3 people from 4 sets of twins if no two
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05 Oct 2014, 01:03
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Thoughtosphere wrote:
How many ways can you group 3 people from 4 sets of twins if no two people from the same set of twins can be chosen?
A. 3 B. 16 C. 28 D. 32 E. 56
As the group shouldn't have siblings in it, then one set of twins can send only one "representative" to the group. The number of ways to choose which 3 sets to send one "representative" to the group is \(C^3_4\) (choosing 3 sets which will be granted the right to send one "representative" to the group);
But each of these 3 sets can send 2 persons to the committee either a brother or a sister: \(2*2*2=2^3\);
Re: How many ways can you group 3 people from 4 sets of twins if no two
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05 Oct 2014, 00:35
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Ways to select 3 people from 8 people (4 twins x 2) = 8C3 = 56 Ways to select 1 twin + 1 people = 4C1*6C1 = 24 Ways to select a group 3 people from 4 sets of twins if no two people from the same set of twins can be chosen = 56 - 24 = 32
Re: How many ways can you group 3 people from 4 sets of twins if no two
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15 Mar 2019, 06:43
UmangMathur wrote:
How many ways can you group 3 people from 4 sets of twins if no two people from the same set of twins can be chosen?
A. 3 B. 16 C. 28 D. 32 E. 56
Let the twins be Aa, Bb, Cc, and Dd (where the uppercase letter denotes the older twin and the lowercase letter the younger twin).
We see that we can group: 1) all 3 uppercase letters, 2) all 3 lowercase letters, 3) 2 uppercase and 1 lowercase letters and 4) 1 uppercase and 2 lowercase letters.
Each of the first 2 options has 4C3 = 4 ways, and each of the last 2 options has 4C2 x 4C1 = 6 x 4 = 24 ways. So the total number of ways is( 2 x 4) + (2 x 24) = 8 + 48 = 56.
Re: How many ways can you group 3 people from 4 sets of twins if no two
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13 May 2019, 10:10
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UmangMathur wrote:
How many ways can you group 3 people from 4 sets of twins if no two people from the same set of twins can be chosen?
A. 3 B. 16 C. 28 D. 32 E. 56
Take the task of selecting the 3 people and break it into stages.
Stage 1: Select the 3 sets of twins from which we will select 1 sibling each. There are 4 sets of twins, and we must select 3 of them. Since the order in which we select the 3 pairs does not matter, this stage can be accomplished in 4C3 ways (4 ways)
Stage 2: Take one of the 3 selected sets of twins and choose 1 person to be in the group. There are 2 siblings to choose from, so this stage can be accomplished in 2 ways.
Stage 3: Take one of the 3 selected sets of twins and choose 1 person to be in the group. There are 2 siblings to choose from, so this stage can be accomplished in 2 ways.
Stage 4: Take one of the 3 selected sets of twins and choose 1 person to be in the group. There are 2 siblings to choose from, 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: D
Cheers, Brent
Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.