probability-pair of twins

[mehrasa]

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?

Ans. 32

I do have difficulty to solve these kind of problem although i read probability sections from OG and Bible.. appreciate if sum1 introduce a great gmat book in probability and permutation section.. thnks

[GMATGuruNY]

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?

Ans. 32

I do have difficulty to solve these kind of problem although i read probability sections from OG and Bible.. appreciate if sum1 introduce a great gmat book in probability and permutation section.. thnks

I posted a solution for a very similar problem -- the same problem, really -- here:

https://www.beatthegmat.com/permutations ... 90334.html

[shankar.ashwin]

Its difficult to explain these sums, let me try..

Here there are a total of 8 people and you have 3 slots of fill.

_ _ _

Try to fill up the slots here,
The first slot can be any one of the 8 people. so I have 8 possibilities.

8 _ _.

The second slot cannot be filled by the person already chosen or his/her sibling , so that leaves us with 6 other people.

So, 8 6 _.

Similarly you could say the 3rd slot can be filled by 4 other people.

So, we have 8 6 4.

Now you gotto remember that of the 3 people chosen it doesnt matter who was chosen first/second or third. (Ordering does not matter, since you can get the same 3 people chosen in different order).

So to remove the ordering among themselves, you gotto divide this by 3!.

So (8*6*4)/3! = 32.

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?

Ans. 32

I do have difficulty to solve these kind of problem although i read probability sections from OG and Bible.. appreciate if sum1 introduce a great gmat book in probability and permutation section.. thnks

[Brent@GMATPrepNow]

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?

Ans. 32

I do have difficulty to solve these kind of problem although i read probability sections from OG and Bible.. appreciate if sum1 introduce a great gmat book in probability and permutation section.. thnks

Take the task of selecting the 3 committee members and break it into stages.

Stage 1: Select the 3 twin pairs from which we will select 1 sibling each.
There are 4 pairs of twins, and we must select 3 pairs. Since the order in which we select the 3 pairs does not matter, we can use COMBINATIONS
We can select 3 pairs from 4 pairs in 4C3 ways (4 ways)

Stage 2: Take one of the 3 selected pairs and choose 1 person to be on the committee.
There are 2 people in the twin pair, so this stage can be accomplished in 2 ways.

Stage 3: Take another of the 3 selected pairs and choose 1 person to be on the committee.
There are 2 people in the twin pair, so this stage can be accomplished in 2 ways.

Stage 4: Take the last of the 3 selected pairs and choose 1 person to be on the committee.
There are 2 people in the twin pair, 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

[regor60]

By subtraction.

Pick which group will be the twins. 4 choices.

Now need to pick 1 person from 6, 6 choices.

Total is 4x6= 24

How many ways to pick 3 people from 8?

8!/3!5!= 56

56-24=...