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Permutations Concept.

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by Smriti Shashikumar » Sun Feb 17, 2013 8:22 am
Kevin needs to form a committee of 3 from a group of 7 engineers for a study improvement of a desing. If 2 of them are too inexperienced to serve together in a group, how many different groups can Kevin form?

A. 10
B. 20
C. 30
D. 35
E. 210

Thanks in advance for the help! Cheers! :)
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by Jim@StratusPrep » Sun Feb 17, 2013 8:39 am
First, figure out the total number of groups:

7*6*5/3! = 35

Then subtract the number of groups that have the 2 inexperienced members:

1*1*5= 5


The answer is 30.
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by Smriti Shashikumar » Sun Feb 17, 2013 9:50 pm
Jim@StratusPrep wrote:First, figure out the total number of groups:

7*6*5/3! = 35

Then subtract the number of groups that have the 2 inexperienced members:

1*1*5= 5


The answer is 30.

How did you calculate the number of groups that have 2 inexperienced members, as in what was the formula you have used?
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by Jim@StratusPrep » Mon Feb 18, 2013 5:25 am
It is not so mucus formula, but rather using logic. The two member have to be on the team, hence 1*1. Then, there are 5 other memebership to make different teams.
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by shenoydevika » Tue Feb 19, 2013 12:01 am
I still don't get it!!!
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