GMAT Practice Question: A certain university will select 1 of 7 candidates eligible to fill a position in the...
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.
\frac{7!}{1!6!} = \frac{7\times\cancel{6!}}{1\times\cancel{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.
\frac{10!}{2!8!} = \frac{10\times9\times\cancel{8!}}{2\times1\times\cancel{8!}} = \frac{10\times9}{2\times1} = 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 \times 45 = 315
Answer
315
