I have tried a couple of iterations of this problem and

Question

I have tried a couple of iterations of this problem and still cannot get the answer of 100. Any help?

gmatnub · Answer

you can have a committee of 3 frenchs or 2 frenches + 1 other, or 1 french + 2 others.

1 french + 2 others --> you get 4 choices among the french and 2 among the 6 others
so 6C2 = 15 
15 * 4 = 60

2 frenches + 1 other --> you get 6 choices among the others, and 2 among the 4 frenches
so 4C2 = 6
6 * 6 = 36

3 frenches --> 3 among the 4 frenches
so 4C3 = 4

60 + 36 + 4 = 100

jasonc · Answer

10 ppl
4 teach only french
6 teach only other

need a team of 3 with at least 1 french teacher
so we have to sum the following:
choose 1 from 4 and 2 from 6
choose 2 from 4 and 1 from 6
choose 3 from 4 and 0 from 6
=>
4*15
6*6
4
=>
60+36+4=100

miccarlo · Answer

Makes sense...not sure why I always try to take the hard way around to calculate these combination questions...better to think about the question first before trying calculations. Thanks!

zonk · Answer

Total number of ways to choose 3 members out of 10 people: 10 choose 3 = 120
Total number of ways to choose 3 members that do not include a french teacher: 6 choose 3 = 20

Total number of ways to choose 3 members with at least 1 french teacher = 120 - 20 = 100

kman · Answer

Could you please tell me what I am messing up here:

For the first spot, we can pick from 4 choices. For the second spot, we can pick from the remaining 9 choices.

I did 4C1 X 9C2 = 116

Evidently this is incorrect, but I can't figure why

I have tried a couple of iterations of this problem and

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