I believe the answer is E (i.e.100)

The wording states that :
- you have 10 members, out of which 4 are French teachers and 6 are either teaching German or Spanish.
- we need to build a 3-member committee with at least 1 French teacher.


Therefore you can have committees with 1, 2 or 3 French teachers (the sum of all those possibitilies being the overall number of committees you're looking for).

number of committees with one French teacher:
you need to choose one teacher out of 4 and two teachers out of 6
i.e. 4C1 * 6C2

number of committees with 2 French teachers:
you need to choose 2 teachers out of 4 and 1 teachers out of 6
i.e. 4C2 * 6C1

number of committees with 3 French teachers:
you need to choose 3 teachers out of 4 and zero teacher out of 6
i.e. 4C3 * 6C0


total committees = 4C1*6C2 + 4C2*6C1 + 4C3*6C0
total committees = 4*15 + 6*6 + 4*1
total committees = 100

szapiszapo wrote:I believe the answer is E (i.e.100)

total committees = 4C1*6C2 + 4C2*6C1 + 4C3*6C0
total committees = 4*15 + 6*6 + 4*1
total committees = 100

Hi,
Could someone explain to me the progression of this in better detail. i.e. how does 4C1*6C2 = 4*15

thanks

somail wrote:

szapiszapo wrote:I believe the answer is E (i.e.100)

total committees = 4C1*6C2 + 4C2*6C1 + 4C3*6C0
total committees = 4*15 + 6*6 + 4*1
total committees = 100

Hi,
Could someone explain to me the progression of this in better detail. i.e. how does 4C1*6C2 = 4*15

thanks

nCp is the writing convention some people use on the forum to express combination (in probability). I just use it too.

given that nCp = n! / p!(n-p)!

4C1*6C2 = 4!/3! * 6!/(2!*4!) = 4*15

same reasoning for the other combinations