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!
Permutations Concept.
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- Jim@StratusPrep
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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.
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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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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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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