"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".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 ?
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Ans E
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! .Geva@MasterGMAT wrote:"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".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
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.
Still couldn't getGeva@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
