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
Can you please solve this for me?...Do explain with steps...
Thanks.
Probability...Help Needed!!!
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- givemeanid
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For math dept, the candidate can be chosen in 7C1 different ways.
For comp dept, the two candidates can be chosen in 10C2 different ways.
So, together, the number of ways 3 candidates can be chosen = 7C1*10C2 = 7*45 = 315
For comp dept, the two candidates can be chosen in 10C2 different ways.
So, together, the number of ways 3 candidates can be chosen = 7C1*10C2 = 7*45 = 315
So It Goes