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

## combinations

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Here's one approach.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

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

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

- faraz_jeddah
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How did you come up with 4*6C1?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 twinsNOTfrom same group ( 56- 24) = 32 ways

- faraz_jeddah
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Hi Brent,[email protected] wrote:Here's one approach.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

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

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.

Thanks many.

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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.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.

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

- faraz_jeddah
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Gotcha![email protected] wrote: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.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.

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

Gracias!!

[email protected] wrote: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.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.

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

Hi Brent,

Thank you for your explanation. Just a small question since I tried to answer it in another way. If I had to calculate the

**number of ways in which the two twins are from same group**, how would I have gone about doing so?

Thanks

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Sure thing.farsar wrote: Hi Brent,

Thank you for your explanation. Just a small question since I tried to answer it in another way. If I had to calculate thenumber of ways in which the two twins are from same group, how would I have gone about doing so?

Thanks

Number of outcomes that OBEY the restriction = (total # of outcomes that IGNORE the restriction) - (# of outcomes that BREAK the restriction)

total # of outcomes that IGNORE the restriction

Select any 3 people from the 8 people

We can do this in 8C3 ways (=

**56**ways)

# of outcomes that BREAK the restriction

So, we want the number of groups that have a pair of twins

Stage 1: Choose a pair of twins to be placed in the group

There are 4 sets of twins. So, we can complete this stage in 4 ways

Stage 2: Choose one more person to be placed in the group (to join the twins selected in stage 1)

There are 6 people remaining, so we can complete this stage in 6 ways

Total number of outcomes that BREAK the restriction = (4)(6) =

**24**

We get:

Number of outcomes that OBEY the restriction = (total # of outcomes that IGNORE the restriction) - (# of outcomes that BREAK the restriction)

=

**56**-

**24**

= 32

Answer: D

Cheers,

Brent

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The portion in red is incorrect: it should be not 4C1 but 2C1.[email protected] wrote:Let the twins be Aa, Bb, Cc, and Dd (where the uppercase letter denotes the older twin and the lowercase letter the younger twin).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

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

If we select two upper case letters from two sets of twins, we must then select one lower case letter from the OTHER TWO SETS OF TWINS, yielding only two options for the lower case letter.

For example:

If we select A and B for the two upper case letters, we must then select c or d for the lower case letter, yielding only two options for the lower case letter.

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