smanstar wrote:Total no of ways of selecting 3 people from a group of 8 people = 8C3 ways = 56 ways

number of ways in which the two twins are from same group is 4 * 6C1 = 24 ways

so no of ways in which the two twins NOT from same group ( 56- 24) = 32 ways

Brent@GMATPrepNow wrote:

sakshis 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

Here's one approach.

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

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775

Brent@GMATPrepNow wrote:

faraz_jeddah wrote: Hi Brent,

Can you give us a more detailed explanation? There are 4 sets of twins but you repeatedly say that we have to select from 3. I think I am missing something very basic in your explanation.

If we select three people, and no two people can be from the same set of twins, then we can think of it from selecting 1 representative from 3 of the 4 twins.

So, stage one of my solution is to first identify the 3 sets of twins from which I will be selecting a representative from each.

For example, let's say the fours sets of twins are A, B, C and D
In stage 1 we select 3 of the sets.
So, let's say we select groups A, B and D

Now that stage 1 has been accomplished (in 4 ways), stage 2 is to take one of the groups (say group A) and select 1 representative.
There are two people in this group, so this can be accomplished in 2 ways.

Stage 3 is to take another group (say group B) and select 1 representative.
There are two people in this group, so this can be accomplished in 2 ways.

Stage 4 is to take the last group (group D) and select 1 representative.
There are two people in this group, so this can be accomplished in 2 ways.

I hope that helps.

Cheers,
Brent

Brent@GMATPrepNow wrote:

faraz_jeddah wrote: Hi Brent,

Can you give us a more detailed explanation? There are 4 sets of twins but you repeatedly say that we have to select from 3. I think I am missing something very basic in your explanation.

If we select three people, and no two people can be from the same set of twins, then we can think of it from selecting 1 representative from 3 of the 4 twins.

So, stage one of my solution is to first identify the 3 sets of twins from which I will be selecting a representative from each.

For example, let's say the fours sets of twins are A, B, C and D
In stage 1 we select 3 of the sets.
So, let's say we select groups A, B and D

Now that stage 1 has been accomplished (in 4 ways), stage 2 is to take one of the groups (say group A) and select 1 representative.
There are two people in this group, so this can be accomplished in 2 ways.

Stage 3 is to take another group (say group B) and select 1 representative.
There are two people in this group, so this can be accomplished in 2 ways.

Stage 4 is to take the last group (group D) and select 1 representative.
There are two people in this group, so this can be accomplished in 2 ways.

I hope that helps.

Cheers,
Brent

Scott@TargetTestPrep wrote:

sakshis 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 last 2 options has 4C2 x 4C1