combinations

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combinations

by sakshis » Mon Sep 24, 2012 9:38 am
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

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by [email protected] » Mon Sep 24, 2012 1:39 pm
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

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by sakshis » Tue Sep 25, 2012 12:56 am
Thank you Brent for your explanation.

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by smanstar » Tue Sep 25, 2012 9:05 pm
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

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by faraz_jeddah » Sun May 19, 2013 12:18 pm
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
How did you come up with 4*6C1?

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by faraz_jeddah » Tue May 21, 2013 1:26 am
[email protected] 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
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.

Thanks many.

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by [email protected] » Tue May 21, 2013 6:57 am
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
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by faraz_jeddah » Tue May 21, 2013 8:22 am
[email protected] 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
Gotcha!
Gracias!!

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by farsar » Wed Mar 13, 2019 12:11 am
[email protected] 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


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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by [email protected] » Wed Mar 13, 2019 9:59 am
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 the number of ways in which the two twins are from same group, how would I have gone about doing so?
Thanks
Sure thing.

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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twins

by GMATGuruNY » Fri Mar 15, 2019 5:42 am
[email protected] 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
The portion in red is incorrect: it should be not 4C1 but 2C1.
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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