Permutations Concept.

[Smriti Shashikumar]

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!

[Jim@StratusPrep]

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.

[Smriti Shashikumar]

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?

[Jim@StratusPrep]

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.

[shenoydevika]

I still don't get it!!!