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permutation

by Manjareev » Sat Sep 22, 2012 8:33 am
A committee of 3 people is to be chosen from four married couples.
what is the number of different committees that can be chosen if two people who are
married to each other cannot both serve on the committee?

16
24
26
30
32

Pls explain the answer.

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by Brent@GMATPrepNow » Sat Sep 22, 2012 9:13 pm
Manjareev wrote:A committee of 3 people is to be chosen from four married couples.
what is the number of different committees that can be chosen if two people who are
married to each other cannot both serve on the committee?

16
24
26
30
32

Pls explain the answer.
Here's one approach.

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

Stage 1: Select the 3 couples from which we will select 1 spouse each.
There are 4 couples, and we must select 3 of them. Since the order in which we select the 3 couples does not matter, this stage can be accomplished in 4C3 ways (4 ways)

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

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

Stage 4: Take one of the 3 selected couples and choose 1 person to be on the committee.
There are 2 people in the couple, 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 = E

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Brent

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by Manjareev » Sat Sep 22, 2012 11:12 pm
Thank you so much Sir for ur reply. ur explanation is clear. i was just confused for the stage 1 which u had mentioned in the answer. now i know my error of reasoning. thanks.

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by LalaB » Mon Oct 01, 2012 8:58 am
four married couples=4*2= 8 people

we need to get 6 ppl(excluding 2 married) out of 8
8!/6!/2!=32
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by Kaustubhk » Wed May 24, 2017 8:51 am
HI brent,

1st member is selected in 8ways

2nd member in 6 ways (we can't select the spouse)

3rd Member in 4 ways

8*6*4
The order in which the is not important so we must divide by the factorial of the number of choices to eliminate over-counting:


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

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by Brent@GMATPrepNow » Wed May 24, 2017 9:14 am
Kaustubhk wrote:HI brent,

1st member is selected in 8ways

2nd member in 6 ways (we can't select the spouse)

3rd Member in 4 ways

8*6*4
The order in which the is not important so we must divide by the factorial of the number of choices to eliminate over-counting:


(8*6*4)/3! = 8*4 = 32
Perfect!

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by Scott@TargetTestPrep » Thu Jun 01, 2017 4:05 pm
Manjareev wrote:A committee of 3 people is to be chosen from four married couples.
what is the number of different committees that can be chosen if two people who are
married to each other cannot both serve on the committee?

16
24
26
30
32
We are given that there are four married couples (or 8 people) and we need to determine the number of ways of choosing 3 people in which no two people are a married couple. This is a special combination problem. Before we tackle this problem, let's review a combination problem with no restrictions.

With no restrictions, the number of ways of choosing 3 people from 8 people is 8C3, which is calculated as follows:

8C3 = 8!/[3!(8-3)!] = 8!/[3!5!] = (8 x 7 x 6)/3! = 56

8, 7, and 6, in the numerator, represent the number of ways the first, second, and third person can be chosen, respectively. We divide the numerator by 3! because in a combination problem we do not care about the order in which the 3 people are chosen.

However, in this (special combination) problem, 3 people are chosen in which no married couple can serve together on the committee. The first person could be any one of the 8 people. However, once a person is selected, that person's spouse cannot also be selected for the committee. This reduces the choices for the second person to 6 possible people (one person has already been selected and that person's spouse now cannot be selected). Once the second person is chosen for the committee, that person's spouse cannot be chosen. This reduces the number of people who could be chosen as the third person to 4. Therefore, the number of ways of choosing these 3 people is:

(8 x 6 x 4)/3! = 32

Thus, there are 32 ways to choose such a committee.

Answer: E

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by regor60 » Fri Oct 12, 2018 8:19 am
Scott@TargetTestPrep wrote:
Manjareev wrote:A committee of 3 people is to be chosen from four married couples.
what is the number of different committees that can be chosen if two people who are
married to each other cannot both serve on the committee?

16
24
26
30
32


With no restrictions, the number of ways of choosing 3 people from 8 people is 8C3, which is calculated as follows:

8C3 = 8!/[3!(8-3)!] = 8!/[3!5!] = (8 x 7 x 6)/3! = 56

8, 7, and 6, in the numerator, represent the number of ways the first, second, and third person can be chosen, respectively. We divide the numerator by 3! because in a combination problem we do not care about the order in which the 3 people are chosen.

Answer: E
Another view, starting with the above is to now recognize that without the restriction in place, there are among the 56 groups of 3 people a number of groups where a couple is represented and the third person. These need to be subtracted from 56 to comply with the restriction.

So how many couples are there ? 4. After accounting for the 2 people in the couple, there are now 8-2 = 6 people left to be the third person. So there are 4x 6 =24 groups of three where a couple is represented. So the correct overall number is then 56-24 = 32