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If a committee of 3 people is to be selected from among 5 ma [#permalink]
25 Jan 2008, 10:36

1

This post was BOOKMARKED

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A

B

C

D

E

Difficulty:

25% (low)

Question Stats:

68% (02:12) correct
31% (00:56) wrong based on 54 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?

(1) ->total combinations = 10c3 = 120 (2) ->total combinations which include atleast one couple = 5*8 = 40 (one couple can be choosen among the 5 couples in 5 ways.. the remaining one slot can be filled with any one of the 8 ppl remaining in 8 ways )

total combinations which do not include any couple = (1) - (2) = 80

Answer is D

another approach is total combinations = (total permutations)/n!

If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include 2 people who are married to each other, how many such committees are possible?

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

First there are 10 people. and we want a committee of 3. So:

10!/3!7! --> 120 ways to have a comittee w/ no constraints.

120 - How many ways can we have couples?

lets say couple one is AA two BB three CC four DD and five EE When we have a couple chosen, we have 8 others left so 8 different possible committees per couple

AAX (X stands for the other 8 members) BBX CCX DDX EEX

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

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

D

Can some explain how to solve this???

Total number of ways to choose 3 people from 10: 10 choose 3 = 10!/(7!*3!) = 120 Total number of ways to include a married couple in the committee = 5*(8 choose 1) = 5*8 = 40

Total number of ways to not include a married couple = 120-40 = 80

where Total number of ways of choosing 3 out of 5 couples = 5C3 = 10 From each couple you can choose only one person. This can be done in 2 ways per couple. For 3 couples, 2^3 = 8 ways

In the explanation Total number of ways to choose 3 people from 10: 10 choose 3 = 10!/(7!*3!) = 120 Total number of ways to include a married couple in the committee = 5*(8 choose 1) = 5*8 = 40

Total number of ways to not include a married couple = 120-40 = 80

How do you get?

Total number of ways to include a married couple in the committee = 5*(8 choose 1) = 5*8 = 40

In the explanation Total number of ways to choose 3 people from 10: 10 choose 3 = 10!/(7!*3!) = 120 Total number of ways to include a married couple in the committee = 5*(8 choose 1) = 5*8 = 40

Total number of ways to not include a married couple = 120-40 = 80

How do you get?

Total number of ways to include a married couple in the committee = 5*(8 choose 1) = 5*8 = 40

If you're going to form a committee with 1 couple and another person, then you have 5 couples to choose from. There are 8 possibilities for the non-couple person from which you choose 1.

cant this question be seen in this way (i know the answer comes wrong ) :

first select any person out of the 10 .

then we have 8 people to select the 3rd person from (excluding the pair above)

then for the 3rd person we have 6 choices (excluding the 2 pairs above)

ans = 10*8*6

(please lemme know why is this wrong ? )

This isn't wrong, it's just incomplete.

You now have your Commitee of 3 people. But you've selected them in an order, when order doesn't matter.

Selection #1 = Jack Selection #2 = Susan Selection #3 = Mary

Assume this is a good committee where Jack is not married to Susan or Mary and Susan and Mary are not married. The above commitee will be the same as the following committee:

Selection #1 = Susan Selection #2 = Mary Selection #3 = Jack

To account for the fact that the order does not matter, we must divide out the number of ways a single committee can be the same members but in different order. 3 places = 3!, 3*2*1 = 6.

10*8*6 / 6 = 10*8 = 80. Answer D.
_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Total possible combination = 8C3 = 56 Total Possible outcome which contain a couple : 4C1*6 (4 combination of couple plus one person from the rest) = 4*6 = 24

Total possible combination = 8C3 = 56 Total Possible outcome which contain a couple : 4C1*6 (4 combination of couple plus one person from the rest) = 4*6 = 24

Ans = 56-24 = 32

In all such problems we need to calculate the total possible combination. therefore as it was calculated above = 56.

We need to choose 3 people from 4 married couples, it will be the following scenarios.

a) 1 married couple and 1 single individual therefore if you selected one couple then 6 people are fighting for the last seat of the committee.

Possible outcome = 4C1 * 6 = 4*6 = 24

Number of possible committees = total - possible = 56 - 24 = 32

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If a committee of 3 people is to be selected from among 5 ma [#permalink]
08 Dec 2013, 06:10

Expert's post

netcaesar 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?

Re: If a committee of 3 people is to be selected from among 5 ma [#permalink]
08 Dec 2013, 06:11

Expert's post

netcaesar 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) each to send one "representative" to the committee: 5C3=10.

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