Aishyk97 wrote:
jet1445 wrote:
In a certain group of 10 members, 4 members teach only French and the rest teach only Spanish or German. If the group is to choose 3-member committee, which must have at least 1 member who teaches French, how many different committee can be chosen ?
A. 40
B. 50
C. 64
D. 80
E. 100
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.