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

Question

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

A. 40
B. 50
C. 64
D. 80
E. 100

Sumithra · Accepted Answer

Total possible combinations = 10C3=120
No French(only Spanish or German)=6c3=20
Atleast one French=120-20=100

amorpheus · Answer

Case 1

1french 2either German or Spanish: 4C1 * 6C2 = 4 * 15 = 60

Case 2

2french 1either German or Spanish: 4C2 * 6C1 = 6 * 6 = 36

Case 3

3french 4C3 = 4

Answer is E = 100

maray · Answer

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

abhimahna · Answer

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

ScottTargetTestPrep · Answer

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

GmatPoint · Answer

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.

Aishyk97 · Answer

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?

Flower_Child · Answer

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.

TheLegacy · Answer

Whenever I see 'at least, at most', I immediately think about finding the total amount of possibilities and subtracting the extreme case.

Total number of possibilities: (10 x 9 x 8) if the order is important, since it is not, we divide by 3 x 2 (3! since there are three spots). Hence total number of possibilities is = to 120.

Now, it is too much, since we are considering everything but we have to subtract the cases in which we do not have any french (since we want the exact opposite, that there is 'at least' one of them).

So, 6 x 5 x 4 divided by 3 x 2 = 20

120 - 20 = 100.

Answer choice (E).

KyleKouton · Answer

Can someone explain why cant we solve by using this logic:

4 (any of the 4 french speakers) X 9C2 (a committee of two people from the rest of 9 members)

I know that by this logic I will get a number which si even bigger than 10C3 but I am not able to intuit what went wrong here

Bunuel · Answer

If a committee has 2 French teachers, say F1 and F2 plus S, then:

You count it once when you choose F1 in the “4” and {F2, S} in 9C2 
 And again when you choose F2 in the “4” and {F1, S} in 9C2

So any committee with k French teachers is counted k times. That overcounting is why 4 * 9C2 does not work.

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

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