How many ways can you group 3 people from 4 sets of twins if no two

Question

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

Bunuel · Accepted Answer

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 ; So total # of ways is C^3_4*2^3=32 . Answer: D. Similar questions to practice: in-a-room-filled-with-7-people-4-people-have-exactly-87550-20.html a-certain-junior-class-has-1-000-students-and-a-certain-58914.html a-dog-breeder-currently-has-9-breeding-dogs-6-of-the-dogs-131992.html three-pairs-of-siblings-each-pair-consisting-of-one-girl-136837.html if-there-are-four-distinct-pairs-of-brothers-and-sisters-99992.html

pqhai · Answer

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

Ans: D

Vetrik · Answer

Twins -> a1,a2; b1b2 ; c1c2 ; d1d2

Choose a1 -> b1 -> C1 ; a1->b1->c2 ; a1 ->b2->c1 ; a1 -> b2 -> c2

Therefore 4 ways.

Now The same can be done with a2 as the first choice . So 4*2 = 8 ways

Now instead of abc....we could choose abd or bcd or acd i.e 4 ways

So 8*4 = 32 ways . Hence Ans D

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Temurkhon · Answer

All case number: 8C3=56

Not desired cases: 4C1*6C1=24

56-24=32

D

ScottTargetTestPrep · Answer

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.

Answer: E

BrentGMATPrepNow · Answer

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.

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satya2029 · Answer

8C3-4C1*6C1
=56-24
=32
D:)

gracex · Answer

I'm not sure if I got the answer by coincidence, but I did this: 
(8 * 6 * 4)/(3 * 2 * 1) = 32

Because for the first slot, there are 8 people to choose from, second slot only 6 people to choose from to avoid siblings, and so on. Then I divided it by 3! because order doesn't matter.

einstein801 · Answer

Hi Brent,

Why is it not just 24*3! ways?

- 24
Fix first guy/girl from couple. second choice -> 6 options. third choice -> 4 options. Total --> 6*4=24

- 3!
Order doesn't matter so multiply by 3!

bumpbot · Answer

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How many ways can you group 3 people from 4 sets of twins if no two

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How many ways can you group 3 people from 4 sets of twins if no two

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