A committee of three people

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A committee of three people

by GMATsid2016 » Mon Dec 05, 2016 8:36 am
A committee of three 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?

A) 16

B) 24

C) 26

D) 30

E) 32

OAE

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by Brent@GMATPrepNow » Mon Dec 05, 2016 8:43 am
GMATsid2016 wrote:A committee of three 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?

A) 16

B) 24

C) 26

D) 30

E) 32

OAE
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)

If anyone is interested, we have a video on calculating combinations (like 4C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

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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by Brent@GMATPrepNow » Mon Dec 05, 2016 8:44 am
GMATsid2016 wrote:A committee of three 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?

A) 16

B) 24

C) 26

D) 30

E) 32

OAE
Another approach is to recognize that: # of permissible committees = total # of 3-person committees that ignore the rule - # of 3-person committees that break the rule

total # of 3-person committees that ignore the rule
If we ignore the rule about married couples, we can select any 3 people from the 8 people.
Since the order of the 3 selected people does not matter, we can use combinations.
We can select 3 people from 8 people in 8C3 ways (= 56 ways)

Aside: If anyone is interested, we have a free video on calculating combinations (like 8C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

# of 3-person committees that break the rule
We want the number of committees consisting of an entire couple and a third person.
Let's take the task of building a 3-person committee and break it into stages.

Stage 1: Select 1 of the 4 couples.
We'll place both people in this couple on the committee.
There are 4 couples, so this stage can be accomplished in 4 ways

Stage 2: Select the third person for the committee
There are now 6 people remaining, so this stage can be accomplished in 6 ways.

By the Fundamental Counting Principle (FCP) we can complete both stages (and thus create a 3-person committee) in (4)(6) ways
In other words, we can create 24 committees that break the rule.

So, # of permissible committees = 56 - 24
= 32
= E

Cheers,
Brent
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by [email protected] » Mon Dec 05, 2016 10:06 am
Hi GMATsid2016,

This question is fairly high-concept, but the math behind it isn't too complex. You have to be organized and you have to remember that we're forming GROUPS of people, so 'duplicate entries' are not allowed.

From the prompt, we have 4 married couples, but we have to form a group of 3 without putting a married couple in the group. Let's work through the logic in pieces...

The 1st person in the group can be ANY of the 8 people. Once we pick one of those 8, then we CANNOT pick the married partner to that person...

For the 2nd person in the group, we now have 6 people to choose from. Once we pick one of those 6, then we CANNOT pick the married partner to that person...

For the 3rd person in the group, we now have 4 people to choose from.

We now multiply those three numbers together: (8)(6)(4) = 192

We're NOT done though. Remember that we're forming GROUPS of people, and the 192 options we've figured out so far contain LOTS of duplicates... If you have 3 people: A, B and C, then that is just ONE group. However, in the above approach, you can end up with that group in 6 different ways:

ABC
ACB
BAC
BCA
CAB
CBA

Thus, every actual option has been counted 6 times, so we have to divide that total by 6...

192/6 = 32

Final Answer: E

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by Matt@VeritasPrep » Thu Dec 08, 2016 8:12 pm
Worst case, look for something close to the obvious wrong answer.

If we're choosing three people from a group of 8, we have 8! / 3!5! options, or 56.

Our answer has to be greater than half of that, so we should have D or E. (If you can't go any further under test conditions, that's OK: you've got a 50% chance of guessing correctly!)

If we wanted to go further, though, let's eliminate our bad cases.

Suppose have the married couple A and B. (A power couple, clearly, with those names!) If they appear in a group of 3, they could appear with any of SIX other people, so there are SIX bad combos with them.

Since there are four married couples in all, there are 4 * 6 = 24 bad combos.

That means our answer = Total Combos - Bad Combos = 56 - 24 = 32.

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by Mseemab » Mon Apr 24, 2017 3:55 am
Hi experts. Can anyone please explain why the below approach is incorrect?

Divide selection of members in three stages for every member.
First member can be chosen in 8 ways.
Second member can be chosen in 6 ways (minus the spouse of first selected member)
Third member can be selected in 4 ways.
Overall, 8*6*4 = 192 ways by which three members can be selected.

Somebody please explain, I am really having a hard time understanding the concepts of permutations and combination. [/list]

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by Jay@ManhattanReview » Mon Apr 24, 2017 4:30 am
Mseemab wrote:Hi experts. Can anyone please explain why the below approach is incorrect?

Divide selection of members in three stages for every member.
First member can be chosen in 8 ways.
Second member can be chosen in 6 ways (minus the spouse of first selected member)
Third member can be selected in 4 ways.
Overall, 8*6*4 = 192 ways by which three members can be selected.

Somebody please explain, I am really having a hard time understanding the concepts of permutations and combination. [/list]
Hi Mseemab,

You can certainly do this; however, the 192 ways gives the answer of how three members can be selected and ARRANGED. We are not interested in the arrangement of the three members.

Say, the three members among the eight are X, Y and Z, and the 192 selections would include XYZ, YXZ, YZX, XZY, ZXY, ZYX; but these are essentially the same selections, so the arrangement of 3 members would be done in 3! = 6 ways. We must divide 192 by 6 to get ONLY selections.

Thus, the answer is [spoiler]912/6 = 32[/spoiler].

Hope this helps!

Relevant book: Manhattan Review GMAT Combinatorics and Probability Guide

-Jay
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by Jeff@TargetTestPrep » Fri Apr 28, 2017 2:00 pm
GMATsid2016 wrote:A committee of three 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?

A) 16

B) 24

C) 26

D) 30

E) 32
We are given that there are 4 married couples (or 8 people), and we need to determine the number of ways of choosing 3 people in which no 2 people are a married couple. So 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 is 8C3, which is calculated as follows:

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

8, 7, and 6 in the numerator represent the number of ways in which 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 choice of 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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