In a certain group of 10 members, 4 members teach only French and the

please correct me if I am right( I am new to this web site)
1. I count all possibilities without restriction: 10!/3!7!=120
2. Count is there is no French members in the committee 6!/3!3!=20
3. 120-20=100

Yes, this is another way to solve this question. I also solved in similar way.

Solution:

Without any restrictions, the number of ways to choose 3 people from 10 is 10C3 = (10 x 9 x 8) / (3 x 2) = 720/6 = 120. Let’s assume a committee can be picked without any member who teaches French; then there are 6C3 = (6 x 5 x 4) / (3 x 2) = 120/6 = 20 ways. So there are a total of 120 different committees (if there are no restrictions) and 20 of them consist of no members who can teach French. Therefore, there must be 120 - 20 = 100 different committees with at least one member who teaches French.

Answer: E

The number of ways 3 professor committees can set up = 10C3 = 10*9*8/3*2 = 120

The number of ways 3 professor committees can be set up without French teachers = 6C3(Since 6 are non-French teachers) = 6*5*4/3*2*1 = 20

Thus, the total number of committees with at least one math professor = 120 - 20 = 100.

Thus, the correct option is E.

Hi,

Can someone explain my gap in understanding with this question?

I thought that since at least 1 member who teaches French should be chosen. 4C1 - So 1 member is chosen and then from 9 I need to choose 2. So I did 9C2.

4*9C2


I understand the explanation given. But where am I going wrong in my approach?

This was an interesting approach, but in this we run into problem of creating duplicate combinations.
Consider that the 4 French teachers are f1 , f2, f3, f4 .
You select one via 4c1. Say f1
now you select 2 from remaining ones. consider that the two people selected are also French teachers say f2, f3
Now we have our 3 teachers - f1, f2 , f3
All okay till here.

Now lets consider one more case. We select one French teacher via 4C1 . This time f2.
From remaining we again select two French teachers - say f1 , f3 ( Since they are in the 9 people group that are left )
so the 3 teachers we have are - f1, f2, f3
Now this is incorrect since we have already selected this combination above.

Thus to avoid this , we have to consider multiple cases in such questions wherever ATLEAST is mentioned.
there are two approaches-
1. To calculate and add all required cases -> 1 French 2 others (4C1 * 6C2) , two French 1 other (4C2 * 6C1) and 3 French (4C3)
2. Find the opposite probability (no French) and minus that from total possible case -> 10C3 - 6C3

Hope this helps.