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by apuso » Wed Jan 20, 2010 8:18 am
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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by sumanr84 » Wed Jan 20, 2010 9:36 am
I think it should be 80.

3 people can be selected in 10C3 ways = 120

By way of reverse calculation and later subtraction we can achieve the result,
Suppose committee consists of 1 couple and one other person.Total No of ways of achieving this = 5C1 * 8C1

Substracting,

10C3 -5C1 8C1 = 80
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by chix475ntu » Wed Jan 20, 2010 1:11 pm
1st person can be selected - 10
2nd person can be selected - 8(not taking the first person and spouse of 1st person)
3rd person can be selected - 6(not taking the first/second person and spouses of 1st/2nd person)

the 3 people can be picked up in 6 different ways and as the order doesnt matter
Ans = 10*8*6/6 = 80
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by VikingWarrior » Sun Jan 24, 2010 9:50 am
the 3 people can be picked up in 6 different ways and as the order doesnt matter
Ans = 10*8*6/6 = 80
Chix I didn't understand how you got this last step...On what basis are you dividing by 6?
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by ajith » Sun Jan 24, 2010 10:03 am
VikingWarrior wrote:
Chix I didn't understand how you got this last step...On what basis are you dividing by 6?
He was changing the permutation to combination

ie. say ABC was one of the combination of people. We have counted BCA , ACB, BAC, CBA and CAB (total six) in the 10*8*6 but all these refer to the same - a combination of person A, Person B and Person C. To nullify the multiple counting he is dividing it by 6 (which is the number of times each combination repeated in the counting)
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by VikingWarrior » Sun Jan 24, 2010 10:33 am
He was changing the permutation to combination

ie. say ABC was one of the combination of people. We have counted BCA , ACB, BAC, CBA and CAB (total six) in the 10*8*6 but all these refer to the same - a combination of person A, Person B and Person C. To nullify the multiple counting he is dividing it by 6 (which is the number of times each combination repeated in the counting)
Thanks for the explanation but I suppose I am number blind so I still have a doubt...if you want to exclude the additional combinations that are counted then shouldn't you be subtracting something from the 10*8*6...I still don't get how you can divide and why by 6?
Hoping for a clear explanation.
Thanks :)[/quote]
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by Brent@GMATPrepNow » Sun Jan 24, 2010 10:45 am
VikingWarrior wrote:
He was changing the permutation to combination

ie. say ABC was one of the combination of people. We have counted BCA , ACB, BAC, CBA and CAB (total six) in the 10*8*6 but all these refer to the same - a combination of person A, Person B and Person C. To nullify the multiple counting he is dividing it by 6 (which is the number of times each combination repeated in the counting)
Thanks for the explanation but I suppose I am number blind so I still have a doubt...if you want to exclude the additional combinations that are counted then shouldn't you be subtracting something from the 10*8*6...I still don't get how you can divide and why by 6?
Hoping for a clear explanation.
Thanks :)
Chix began by treating this question as an arrangement question. In other word, order matters.
So, selecting persons A, B, and E is different from selecting E, B and A

Of course, order doesn't matter here (selecting persons A, B, and E is is the same as selecting E, B and A), so within the 10x8x6 arrangements calculated, each unique arrangement is counted more than once. How many times is each unique arrangement counted?

Let's examine the selection A, B, and E.
If we say that order matters, then we will treat the following arrangements as unique:
A, B, E
A, E, B
B, A, E
B, E, A
E, A, B
E, B, A
However, these arrangements are not unique. These 6 arrangements represent the same selection of 3 people.
So, we can conclude that each unique selection has been counted 6 times.
So, to get our answer, we need to divide 10x8x6 by 6

Aside: We can arrange 3 unique things in 3! (6) ways. Similarly, we can arrange 4 unique things in 4! (24) ways. And, in general, we can arrange n unique things in n! ways.
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by ajith » Sun Jan 24, 2010 10:47 am
VikingWarrior wrote:
Thanks for the explanation but I suppose I am number blind so I still have a doubt...if you want to exclude the additional combinations that are counted then shouldn't you be subtracting something from the 10*8*6...I still don't get how you can divide and why by 6?
Hoping for a clear explanation.
Thanks :)
What if you have counted each combination 6 times. Just as ABC was counted 6 times - all the possible combinations were counted 6 times. Thus we divide by six to get unique combinations
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by VikingWarrior » Sun Jan 24, 2010 11:07 am
Thank you Brent and Ajith...Math was a loong time ago...very well explained :)
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by diebeatsthegmat » Wed Jun 02, 2010 5:58 pm
apuso 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
hi, can i solve this maths by this way.
first of all, i select 3 people in group of 5 married couples = 10 people : 10!/(7!*3!)=120
then i choose the way to select a committee of 3 people which includes 2 married people:
because there are 5 couples so there are 5! to choose married people and we have to select just 3
so 5!/3=1*2*3*4*5/3=40
and 120 -40=80
can i solve the maths by the above method? i just read the way to solve probability and combination of Manhattan and kaplan thus i am not sure if i am correct?!?!
help please...:)
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