Problem Solving

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 a 3-member committee , which must have atleast 1 member who teaches French, how many different committees can be chosen ?

40
50
64
80
100

Ans E

1-french and 2 other language : 4!/1!3! * 6!/2!4! = 4*15= 60
2-french and 1 other langauage : 4!/2!2! * 6!/1!5! =6*6= 36
All 3 french : 4!/3!1! = 4

Add all three outcomes : 60+36+4 = 100

correct ans [E]

Good, straight forward answer.

I guess I need to go back and look at probabilities because I had a lot of trouble with this one.

moneyman 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 a 3-member committee , which must have atleast 1 member who teaches French, how many different committees can be chosen ?

40
50
64
80
100

Ans E

"At least" is a sign to go (total number of combinations) - (number of forbidden combinations). If we want at least one french speaking member, the only forbidden option is "no french speaking members".

So:
total number of possible combinations, without limitations (choose 3 out of 10, regardless of who we choose) = 10!/7!3! = 10*9*8/6 = 5*3*8 = 120
minus
Number of no french forbidden combinations (choose 3 people out of only 6 non-french speakers) = 6!/3!3! = 6*5*4/3! = 5*4 = 20

Result is 120-20 = 100.

Geva@MasterGMAT wrote:

moneyman 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 a 3-member committee , which must have atleast 1 member who teaches French, how many different committees can be chosen ?

40
50
64
80
100

Ans E

"At least" is a sign to go (total number of combinations) - (number of forbidden combinations). If we want at least one french speaking member, the only forbidden option is "no french speaking members".

So:
total number of possible combinations, without limitations (choose 3 out of 10, regardless of who we choose) = 10!/7!3! = 10*9*8/6 = 5*3*8 = 120
minus
Number of no french forbidden combinations (choose 3 people out of only 6 non-french speakers) = 6!/3!3! = 6*5*4/3! = 5*4 = 20

Result is 120-20 = 100.

Could you please explain me this - 10!/7!3! = 10*9*8/6 = 5*3*8 = 120. I dont understand how you get 5*3*8 out of 10!/7!3! .

sorry I skipped a step.

First, reduce the 10! with the 7! = 10*9*8*7*6... / 7*6*... = you're left with 10*9*8.

So 10!/7!3! is first reduced to
10*9*8/3! = 10*9*8 / 3*2*1

Next reduce the 3 with the 9, and the 2 with the 10, to get
5*3*8/1

Geva@MasterGMAT wrote:sorry I skipped a step.

First, reduce the 10! with the 7! = 10*9*8*7*6... / 7*6*... = you're left with 10*9*8.

So 10!/7!3! is first reduced to
10*9*8/3! = 10*9*8 / 3*2*1

Next reduce the 3 with the 9, and the 2 with the 10, to get
5*3*8/1

Still couldn't get

10C3-6C3= 100