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# If a committee of 3 people is to be selected from among 5

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If a committee of 3 people is to be selected from among 5 [#permalink]

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05 Jan 2010, 07:46
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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 committees are possible?

A. 20
B. 40
C. 50
D. 80
E. 120
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Re: committee of 3 [#permalink]

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05 Jan 2010, 10:07
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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 committees are possible?

A. 20
B. 40
C. 50
D. 80
E. 120

Each couple can send only one "representative" to the committee. Let's see in how many ways we can choose 3 couples (as there should be 3 members) to send only one "representatives" to the committee: 5C3=10.

But these 3 couples can send two persons (husband or wife): 2*2*2=2^3=8.

Total # of ways: 5C3*2^3=80.

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Re: committee of 3 [#permalink]

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05 Jan 2010, 09:03
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kirankp 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 committees are possible?
A. 20
B. 40
C. 50
D. 80
E. 120

total no for selecting 3 out of 10=10c3=120

no. of ways in which no two married people included= tot- 2 married couple included
2 married couple can be included in 5c1( no. of ways selecting a couple) * 8c1( no. of ways selecting the third person)=5 * 8=40

reqd comb=120-40=80
hence D
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Re: committee of 3 [#permalink]

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06 Jan 2010, 11:34
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I like to think of it like this:

Step 1 - find the combinations without any restrictions

10C3 = 120

Step 2 - subtract the combinations that would have a couple in the committee

5C1 x 4C1 x 2 = 40

In this step, we first find the # of ways to choose a couple, which is 5C1=5.
After getting the first couple, we need 1 more member, so we choose 1 couple of the remainin 4 couples, which is 4C1 = 4. But within this new couple, we can either choose the man or the woman, so we need to x2.

Step 3 - find answer (no restrictions minus restrictions)

120 - 40 = 80

So the answer is 80.

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Re: committee of 3 [#permalink]

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05 Jan 2010, 08:06
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ANS -80..
total people=10.. ways to select 3 out of them=10c3=120...
it includes comb including couple..
ways in which couple are included =8c1*5=40..
so ans reqd 120-40=80...
(if we take a gp to include a couple ,it will include couple +any one of rest 8 so 8c1 ways ..
5 couple so 5*8c1=40)

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Re: If a committee of 3 people is to be selected from among 5 [#permalink]

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25 Oct 2013, 19:01
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Using slot method:
First person can be chosen -> 10 ways,
2nd person can be chosen -> 8 ways (1st person and his wife are not candidates) and
3rd person can be chosen -> 6 ways (1st person/2nd person and their wives are out)
Answer -> 10*8*6/6 = 80 (divide by 6 because the 3 people can be chosen in any order (i.e. 3! = 3*2*1 ways))

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Re: committee of 3 [#permalink]

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11 Oct 2010, 02:43
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"If a committee of 3 people is to be selected"
Combo box arrangement
(_)(_)(_)/3!

"from among 5 married couples"
Bag of 10 choices: A,B,C,D,E,F,G,H,I,J

"so that the committee does not include two people who are married to each other"
First slot has 10 choices
(10)(_)(_)/3!

but the choice eliminates the spouse. The second slot has 8 choices
(10)(8)(_)/3!

but the choice eliminates another spouse. The third slot has 6 choices
(10)(8)(6)/3!

"how many such committees are possible?"
(10)(8)(6)/(3*2) = 80

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Re: Combinatorics question [#permalink]

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22 May 2014, 22:26
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Hi achakrav2694,

Ordering is not Required in Selection.

We can have just 4 Cases:

1) Selecting All 3 husbands: This would be$$5C3 = 10$$
2) Selecting 2 husbands and 1 Wife: $$5C2 * 3$$ (As Wife cannot be for the 2 husbands selected) $$= 30$$
3) Selecting All 3Wives: This would be $$5C3 = 10$$
4) Selecting 2 Wives and 1 Husband: $$5C2 * 3$$(As Husband cannot be for the 2 Wives selected) $$= 30$$

Total Commitees $$= 10 + 30 + 10 + 30 = 80$$

Rgds,
Rajat
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Re: If a committee of 3 people is to be selected [#permalink]

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23 May 2014, 00:06
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achakrav2694 wrote:
I would like to know the best way to approach a problem similar to the one below, or if anyone has any tricks to solve it. I don't fully understand the GMAC explanation.

Q: If a committee of 3 people is to be selected from 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
A) 20
B) 40
C) 50
D) 80
E) 120

Answer is D, 80.

The explanation given says that there are 10 people who can be the first, 8 people who can be the second and 6 that can be the third. Because there are 6 ways of ordering 3 people, the answer is (10*8*6)/6. I find this slightly confusing and unintuitive. Does anyone have a method or easier explanation?

Hello achakrav2694,

The Question is discussed here : if-a-committee-of-3-people-is-to-be-selected-from-among-88772.html#p669715

Before posting a question, It is worthwhile to use Search option to check if the question has been answered before. Since this is your 2nd post, I would suggest you to go through the below links for better navigation on Gmatclub

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If a committee of 3 people is to be selected from among 5 [#permalink]

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26 Aug 2016, 09:53
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kirankp 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 committees are possible?

A. 20
B. 40
C. 50
D. 80
E. 120

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.

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.

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Last edited by ScottTargetTestPrep on 27 Aug 2016, 06:02, edited 2 times in total.

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Re: committee of 3 [#permalink]

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07 Jan 2010, 05:30
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I too got 80 with the conventional way of 10C3 - 5C1 * 8C1 = 120 - 40 = 80.
But learnt and loved Bunuel's way. Thanks!

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Re: committee of 3 [#permalink]

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15 Jan 2011, 14:55
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srivicool wrote:
can you please explain the combo box arrangement explanation for the problem ??

i am not able to understand how we get 3! in the denominator ??

This issue is discussed here: ps-combinations-94068.html and here: if-there-are-four-distinct-pairs-of-brothers-and-sisters-99992.html
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Re: committee of 3 [#permalink]

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12 Dec 2013, 22:29
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Bunuel, please correct me if i'm wrong. Thank you for your help, i appreciate it!

10*8*6=480 (we chose 3 people out of 10 so that no couple included)
480/3!= 80 (un-arranged the order as it doesn't matter)

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Re: If a committee of 3 people is to be selected from among 5 [#permalink]

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15 Jul 2014, 04:06
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Bunuel 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 committees are possible?

A. 20
B. 40
C. 50
D. 80
E. 120

Each couple can send only one "representative" to the committee. Let's see in how many ways we can choose 3 couples (as there should be 3 members) to send only one "representatives" to the committee: 5C3=10.

But these 3 couples can send two persons (husband or wife): 2*2*2=2^3=8.

Total # of ways: 5C3*2^3=80.

wow superb approach I would give you 100 Kudos. THANKS Bunuel.
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If a committee of 3 people is to be selected from among 5 [#permalink]

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15 Sep 2014, 18:02
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For those as challenged with math as myself, try this (apologies if this has already been done):

Consider the "slot method"

We need to select three people out a total of ten people (5 couples=10ppl), so create three "slots"

_ _ _
1 2 3

The only restrictions we have are that we cant pick an individual and that individual's spouse, nor can we select any individual twice. Knowing this, we can put any of our 10 people in the first slot. The second spot, however, will be limited to 8ppl (10 total less the person we placed in spot one, leaves 9, less that person's spouse, leaves 8). The third slot will only have 6 possibilities (10 total less the two ppl already placed and each of their spouses). Fill in the slots:

10 8 6
1 2 3

We would then multiply across to get 10*8*6=480

When using the Slot method, if the order of the selections does not matter (as is the case here) we must divide the the product of the slots by n!, where n is just the number of slots.

480/3! =480/6 = 80

Generally, you'd want to simplify before dividing, so the 2*3 in the 3! term would cancel the 6 in the numerator, and you'd just be left with (10*8)/1 or 80.

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Re: committee of 3 [#permalink]

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06 Jan 2010, 11:37
Bunuel 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 committees are possible?

Each couple can send only one "representative" to the committee. Let's see in how many ways we can choose 3 couples (as there should be 3 members) to send only one "representatives" to the committee: 5C3=10.

But these 3 couples can send two persons (husband or wife): 2*2*2=2^3=8.

Total # of ways: 5C3*2^3=80.

I actually like this way of thinking more though.

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Re: committee of 3 [#permalink]

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27 Feb 2010, 10:39
I understand the 1-x approach, but if I were to do it the straighforward way, I get

10 x 8 x 6 (first place 10 ways, second place 8 ways, third place 6 ways) = 480, which is wrong.

What am I missing here?
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Re: committee of 3 [#permalink]

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28 Sep 2010, 08:38
Bunuel 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 committees are possible?

Each couple can send only one "representative" to the committee. Let's see in how many ways we can choose 3 couples (as there should be 3 members) to send only one "representatives" to the committee: 5C3=10.

But these 3 couples can send two persons (husband or wife): 2*2*2=2^3=8.

Total # of ways: 5C3*2^3=80.

I like this way of thinking and the calculations seem simpler and quicker.

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Re: committee of 3 [#permalink]

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15 Jan 2011, 04:28
can you please explain the combo box arrangement explanation for the problem ??

i am not able to understand how we get 3! in the denominator ??

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Re: If a committee of 3 people is to be selected from among 5 [#permalink]

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26 Nov 2013, 00:53
I solved the question in different way.

first I computed the number of ways of selecting 3 out of 10, which is 120

second I computed the probability of selecting 3 unmarried people out of 5 couples = 10/10 * 8/9 * 6/8 = 2/3

finally multiplying the total number of selection by the probability of selecting 3 unmarried people 2/3 * 120 = 80
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Re: If a committee of 3 people is to be selected from among 5   [#permalink] 26 Nov 2013, 00:53

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