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by ketkoag » Wed Apr 08, 2009 4:45 am
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

OA: E
Now i got the answer right. i've solved it by (10 * 9 * 7)/2.
please lemme know whether there is other way to solve it. or please lemme know any other example similar to this so that we can clear the concept thru problems.
thanks
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by scoobydooby » Wed Apr 08, 2009 5:30 am
1 candidate for maths can be chosen in 7C1 or 7 ways.
2 candidates for computer sc. can be chosen in 10C2 or 45 ways.

total number of ways in which 3 candidates can be chosen is 45*7=315 ways (independent events)
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by mike22629 » Wed Apr 08, 2009 8:45 am
Not sure if you did this, but the only way to make the calculation easier is to further simply by make the 10 in the numerator 5, and canceling out the 2 in the denominator.
Other than that, your approach is probably the fastest way.
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