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If a committee of 3 people is to be selected from among 5 [#permalink]
05 Jan 2010, 06:46

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Difficulty:

25% (medium)

Question Stats:

71% (01:44) correct
29% (01:12) wrong based on 468 sessions

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?

Re: committee of 3 [#permalink]
05 Jan 2010, 07:06

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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)

Re: committee of 3 [#permalink]
05 Jan 2010, 08:03

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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 _________________

GMAT is not a game for losers , and the moment u decide to appear for it u are no more a loser........ITS A BRAIN GAME

Re: committee of 3 [#permalink]
05 Jan 2010, 09:07

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Expert's post

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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

Another way to think about this problem:

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.

Re: committee of 3 [#permalink]
06 Jan 2010, 10:34

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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)

Re: committee of 3 [#permalink]
06 Jan 2010, 10:37

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?

Another way to think about this problem:

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.

Re: committee of 3 [#permalink]
28 Sep 2010, 07:38

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?

Another way to think about this problem:

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.

Re: If a committee of 3 people is to be selected from among 5 [#permalink]
25 Oct 2013, 18:01

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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))

Re: If a committee of 3 people is to be selected from among 5 [#permalink]
24 Apr 2014, 12:48

saintforlife wrote:

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))

I got this slot method but unfortunately I am unable to get the same answer using the box method.

I first find the number of ways I can find a couple: 10(Can choose any person) * 1(Needs to be the spouse of the person chosen in the first place) * 8(Can be any of the remaining 8) = 80/3!

Total possible combinations = 10C3 = 120

Therefore number of combinations for unmarried couples = 120-(80/3!) which is obviously the wrong answer. Why is this approach wrong ?

If a committee of 3 people is to be selected [#permalink]
22 May 2014, 19:17

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?

Re: Combinatorics question [#permalink]
22 May 2014, 21:26

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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 _________________

If you liked the post, please press the'Kudos' button on the left

Re: If a committee of 3 people is to be selected [#permalink]
22 May 2014, 23:06

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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?

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

Re: If a committee of 3 people is to be selected [#permalink]
23 May 2014, 00:39

Expert's post

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?

Merging similar topics. Please refer to the discussion above.

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