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permutation

This topic has 3 expert replies and 3 member replies
Manjareev Junior | Next Rank: 30 Posts Default Avatar
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permutation

Post 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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Post 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!

Cheers,
Brent

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Post 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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Post 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!

Cheers,
Brent

_________________
Brent Hanneson – Founder of GMATPrepNow.com
Use our video course along with Beat The GMAT's free 60-Day Study Guide

Check out the online reviews of our course
Come see all of our free resources

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Post 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

_________________

Scott Woodbury Stewart Founder & CEO
GMAT Quant Self-Study Course - 500+ lessons 3000+ practice problems 800+ HD solutions
5-Day Free Trial 5-DAY FREE, FULL-ACCESS TRIAL TTP QUANT

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Kaustubhk Junior | Next Rank: 30 Posts Default Avatar
Joined
27 Feb 2015
Posted:
25 messages
Upvotes:
1
Post 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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LalaB Master | Next Rank: 500 Posts
Joined
08 Dec 2010
Posted:
425 messages
Followed by:
7 members
Upvotes:
56
GMAT Score:
690
Post 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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