GMAT Combinations: If none of the candidates is eligible…
Question
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?
- 42
- 70
- 140
- 165
- 315
Topics: combinations, p&c multisource
Solution
Step 1
We first calculate the number of ways to choose one mathematics candidate from seven eligible candidates.
(7!)/(1!6!) = 7×6!1×6! = 7
Step 2
We then calculate the number of ways to choose two computer science candidates from ten eligible candidates, where the two positions are identical.
(10!)/(2!8!) = 10×9×8!2×1×8! = (10×9)/(2×1) = 45
Step 3
Because the mathematics selection and the computer science selection are independent, we multiply these counts to find the total number of sets of three candidates.
7 × 45 = 315
Answer
315 (E)
