If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committee are possible.
A) 20
B) 40
C) 50
D) 80
E) 120
OAD
Please explain.
Many thanks in advance.
Kavin
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If a committee of 3 people is to be selected
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DavidG@VeritasPrep
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If there are 5 couples, there are 10 people to choose from. We can pick any of those 10 for our first position. For our second selection, there are 9 people remaining, but we can't select the first person's spouse. So we have 8 options for our second position. For our third selection, there are 8 people remaining, but we can't pick the spouses of the first two people, leaving us with 6 options.Needgmat wrote:If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committee are possible.
A) 20
B) 40
C) 50
D) 80
E) 120
OAD
Please explain.
Many thanks in advance.
Kavin
First position: 10 options
Second position: 8 options
Third position: 6 options.
But these positions are interchangeable. A group of Mary, Joe, and Amy would be the same as a group of Amy, Mary and Joe. So we need to divide by 3! to account for this.
10*8*6/3! = 80. Answer is D
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dustystormy
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Assume that we have 10 people as
A1A2
B1B2
C1C2
D1D2
E1E2
Now to select 3 people from 10 people is 10C3 = 120. This includes case of 1 couple, which we need to subtract.
chose 1 couple from the 5 as 5C1 (say A1A2)
then from the 4 remaining couple we need to chose 1 person.
So chose 1 couple from the remaining 4 by 4C1 (say B1B2)
Now from B1B2 we need 1 person se chose by 2C1
Finally it becomes 5C1*4C1*2C1 = 40
Therefore total of number of ways of forming a committee without a couple is 120-40 = 80.
I hope it helps.
A1A2
B1B2
C1C2
D1D2
E1E2
Now to select 3 people from 10 people is 10C3 = 120. This includes case of 1 couple, which we need to subtract.
chose 1 couple from the 5 as 5C1 (say A1A2)
then from the 4 remaining couple we need to chose 1 person.
So chose 1 couple from the remaining 4 by 4C1 (say B1B2)
Now from B1B2 we need 1 person se chose by 2C1
Finally it becomes 5C1*4C1*2C1 = 40
Therefore total of number of ways of forming a committee without a couple is 120-40 = 80.
I hope it helps.
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Brent@GMATPrepNow
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Take the task of creating a committee and break it into stages.Needgmat wrote:If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committee are possible.
A) 20
B) 40
C) 50
D) 80
E) 120
Stage 1: Select 3 COUPLES
Since the order in which we select the couples does not matter, we can use COMBINATIONS
We can select 3 couples from 5 couples in 5C3 ways ( = 10 ways)
ASIDE: If anyone is interested, we have a free video on calculating combinations (like 5C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789
At this point, we have 3 COUPLES, which we'll call A, B ans C. We're now going to select ONE person from each couple to be on the committee.
Stage 2: Select 1 person from couple A
There are 2 people in this couple, so we can complete this stage in 2 ways.
Stage 3: Select 1 person from couple B
There are 2 people in this couple, so we can complete this stage in 2 ways.
Stage 4: Select 1 person from couple C
There are 2 people in this couple, so we can complete this stage in 2 ways.
By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus create a 3-person committee) in (10)(2)(2)(2) ways ([spoiler]= 80 ways[/spoiler])
Answer: D
--------------------------
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Jeff@TargetTestPrep
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We are given that there are five married couples (or 10 people) and we need to determine the number of ways of choosing 3 people in which no two 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.Needgmat wrote:If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committee are possible.
A) 20
B) 40
C) 50
D) 80
E) 120
OAD
With no restrictions, the number of ways of choosing 3 people from 10 people is 10C3, which is calculated as follows:
(10 x 9 x 8)/3! = 120
10, 9 and 8, 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 10 people. However, once a person is selected, that person's spouse cannot also be selected for the committee. This reduces the second person to 8 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 6. Therefore, the number of ways of choosing these 3 people is:
(10 x 8 x 6)/3! = 80
Thus, there are 80 ways to choose such a committee.
Answer: D
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Matt@VeritasPrep
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Another approach: choose three distinct couples, then choose exactly one half of each couple. (I think this is a LOT easier.)
Choosing the couples = (5 choose 3) => 5!/3!2! => 10
Choose one of the two members of each couple = (2 choose 1)*(2 choose 1)*(2 choose 1) => 8
So we've got 10 * 8, or 80.
Choosing the couples = (5 choose 3) => 5!/3!2! => 10
Choose one of the two members of each couple = (2 choose 1)*(2 choose 1)*(2 choose 1) => 8
So we've got 10 * 8, or 80.
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DHILLONRAVI1983
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David, I have a basic confusion here. Why is 10C1 * 8C1 * 6C1 wrong?
DavidG@VeritasPrep wrote:If there are 5 couples, there are 10 people to choose from. We can pick any of those 10 for our first position. For our second selection, there are 9 people remaining, but we can't select the first person's spouse. So we have 8 options for our second position. For our third selection, there are 8 people remaining, but we can't pick the spouses of the first two people, leaving us with 6 options.Needgmat wrote:If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committee are possible.
A) 20
B) 40
C) 50
D) 80
E) 120
OAD
Please explain.
Many thanks in advance.
Kavin
First position: 10 options
Second position: 8 options
Third position: 6 options.
But these positions are interchangeable. A group of Mary, Joe, and Amy would be the same as a group of Amy, Mary and Joe. So we need to divide by 3! to account for this.
10*8*6/3! = 80. Answer is D
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DavidG@VeritasPrep
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Because we're not selecting from three separate populations. Think of it this way: if we didn't have the restriction that we don't want a married couple, and we simply wished to select three individuals from the pool of 10, that's a simple combination, so we'd just find 10C3, right?David, I have a basic confusion here. Why is 10C1 * 8C1 * 6C1 wrong?
But 10C3 clearly isn't the same as doing 10C1 * 9C1 * 8C1. The latter would be the correct calculation if Group A contained 10 people, Group B contained 9 people and Group C contained 8 people, and we wished to know how many groups we could form if we selected one person from A, one person from B, and one person from C. In the problem in question, we have a single group with 10 people in it.
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DavidG@VeritasPrep
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It might also be helpful to consider a case in which the numbers are quite small and manageable. Imagine you have two married couples and you wish to make a group of 2 in which the 2 people are not married to each other. Call one married couple Aa and the other Bb. For the first pick, you could select any of the 4 people. For the second pick, because we can't select that person's spouse, there are only 2 options left. Last, because a group of AB is the same as BA, we need to divide by 2! to account for the fact that the two positions in the group are interchangeable, so the answer is 4*2/2! = 4. We can confirm this by listing out the possibilities.
AB
Ab
aB
ab
If we did 4C1*2C1, we'd end up with 8, as we'd be neglecting to divide by 2! (But if we had one group of ABCD and a second group of EF, and we wanted one from each group, then 4C1 * 2C1 would be correct.)
AB
Ab
aB
ab
If we did 4C1*2C1, we'd end up with 8, as we'd be neglecting to divide by 2! (But if we had one group of ABCD and a second group of EF, and we wanted one from each group, then 4C1 * 2C1 would be correct.)
