Ok let's see. The question is:
Quote:
1. A club has 8 male and 8 female members. The club is choosing a commitee of 6 members. The commitee must have at least 2 male and 2 female members. How many different commitees must be chosen ?
Approach 1:
First choose 2 male from the 8 male: C(8,2)
Then choose 2 female from the 8 female: C(8,2)
Then choose the 2 remaining from the 12 remaining: C(12,2)
Total outcome = C(8,2)*C(8,2)*C(12,2) = (8*7/2)*(8*7/2)*(12*11/2) = (8*7/2)*(8*7/2)*(66)
Approach 2:
Total outcome for choosing 2 male and 4 female: C(8,2)*C(8,4)
Total outcome for choosing 3 male and 3 female: C(8,3)*C(8,3)
Total outcome for choosing 4 male and 2 famale: C(8,4)*C(8,2)
Total outcome = C(8,2)*C(8,4)*2+C(8,3)*C(8,3)
= (8*7/2)*(8*7*6*5/4*3*2*1)*2+(8*7*6/3*2*1)*(8*7*6/3*2*1)
= (8*7/2)*(8*7/2)*[(6*5/4*3)*2 + (6/3)*(6/3)]
= (8*7/2)*(8*7/2)*(9)
Compare the answer for the two approaches, obviously 9 is different from 66. What is wrong?
The mistake of approach 1 is that it has included some cases multiple times. For example, say we chose male A and B, female C and D, and then we choose male E and female F from the rest of people. This is counted as one outcome. However, we could also chose male A and E, female C and D, and then we choose male B and female F from the rest of the people. It is counted as another outcome. On a closer look however, you'll notice that these two supposedly different outcomes are actually the same thing.
Normally we'd be able to get to the number of non-ordered outcomes from the number of ordered outcomes by dividing the first by the number of different orders. However in this case it is kind of hard to get the number of different orders. So I would recommend we use approach 2 to solve this question.