Combination Problem(plz help)

akpareek wrote:A certain university will select 1 of 7 candidates eligible to fill a position in the
mathematics department and 2 of 10 candidates eligible to fill 2 identical positions in the
computer science department. If none of the candidates is eligible for a position in both
departments, how many different sets of 3 candidates are there to fill the 3 positions?
A. 42
B. 70
C. 140
D. 165
E. 315

If no one is eligible, 0 candidates will be chosen.. am i missing something here?

If all are eligible, then the required number of sets = 7C1 * 10C2 = 315

7C1 * 10C2 = 315

akpareek wrote:A certain university will select 1 of 7 candidates eligible to fill a position in the
mathematics department and 2 of 10 candidates eligible to fill 2 identical positions in the
computer science department. If none of the candidates is eligible for a position in both
departments, how many different sets of 3 candidates are there to fill the 3 positions?
A. 42
B. 70
C. 140
D. 165
E. 315

The best way to do a problem like this one that involves selecting people from 2 different groups is to calculate each one seperately and then multiply the two results you get.

so.....since this problem involves selecting people from a group, the order is not important and hence this is a combination problem.

so group 1: 7! / 1!6! = 7
group 2: 10!/2!8! = 45

7x45 = 315 and thats your answer.