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Probability

by deepakb » Wed Oct 06, 2010 10:45 pm
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 shovan85 » Wed Oct 06, 2010 10:55 pm
Say there are three spaces A B C

Total number of ways to fill 3! = 6


A can be filled in 10 ways as out of all 5 couples any 1 can be selected

B can be filled in 8 ways Except the selected person and his/her spouse rest 8 ppl left.

and Similarly C can be filled in 6 ways

So the total number of ways = ( 10*8*6 )/ 6 = 80

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by deepakb » Wed Oct 06, 2010 11:22 pm
thanks.
But i didn't understand why you divided by 3!=6.

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by yuliawati » Wed Oct 06, 2010 11:48 pm
You can solve this problem by calculating the total numbers of committe that includes married couple.

Lets say the committee is AAC (AA represents a couple, and C represents the third person in the committe).

1. Total numbers of ways to select the committee = 10C3 (5 couples = 10 person) = 120.

2. There are 5 couples. Total number of ways to choose 3 of 5 couples = 5C3 = 5*4/2 = 10.

3. The third person in the committee can be selected from the other 2 couples --> 4C1 = 4.

4. Multiply #2*#3 you get 10*4 = 40 ways to select the committee that includes married couple.

5. To get the total possible numbers of committee that doesnt include married couple, subtract 40 from the total = 120-40 = 80 (Answer D).

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by shovan85 » Thu Oct 07, 2010 1:05 am
deepakb wrote:thanks.
But i didn't understand why you divided by 3!=6.
Total number of places taken into account is 3.
The number of ways they will be filled is 3! = 6.

ABC
ACB
BCA
BAC
CAB
CBA

But here the order does not matter for us as we have to select the concerned people once irrespective of what place he or she takes. So to remove this we divide the result by 3!. If you do not divide then 10*8*6 will be the result of 6 types of arrangemet among them.

Hope this makes sense :)

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by deepak123gmat » Thu Oct 07, 2010 6:17 am
thanks shovan.

I was just getting comfortable with addition and multiplication in probability now division.
i really dont get when to use all these.

Can anyone share these things? please ............

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by neerajkumar1_1 » Thu Oct 07, 2010 6:40 am
I suppose Shovan has explained it pretty well...

Here is another take on it...

I suppose the first thing u should realize is that
the number of ways of selecting 3 people out of 10 is = 10C3 = 120...

so this is the number of people which we can choose without any limitations..

Now there is one additional limitation provided by the question...
and i.e that no 2 people should be couples...

so all we need to do is subtract those combinations from the total of 120...

now imagine that out of 3 places... the first 2 are occupied by one couple
A1 A2
now this last place can be filled by any of the 8 people left...

so that makes 8 combinations with the first couple...

since there are total of 5 couples... a total of 8 * 5 = 40 combinations will be the ones which we dont want...

Hence the net combinations = 120 - 40 = 80

Pick D

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by deepak123gmat » Thu Oct 07, 2010 6:48 am
Thanks Neeraj, now i got it.
I am not good at probability.