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

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 [or 4C3 ways]

So 8*4 = 32 ways . Hence Ans D

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Please give Kudos if the post helps

All case number: 8C3=56

Not desired cases: 4C1*6C1=24

56-24=32

D

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

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

­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. 

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!
­