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Help! combination/probability

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by silvia928 » Fri May 30, 2008 10:17 am
A certain university will select 1 of 7 candidates eligible to fill a position in the math department and 2 of 10 candidates eligible to fill 2 identical positions in the english 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

Ans. (E)
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by aatech » Fri May 30, 2008 10:41 am
From first group 1 person can be selected in 7 ways

From second group 2 person can be selected in 10C2 = 45 ways

Total no of combo = 7*45 = 315
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Re: Help! combination/probability

by beeparoo » Fri May 30, 2008 1:01 pm
silvia928 wrote:...how many different sets of 3 candidates are there to fill the 3 positions?
I think I have difficulty interpreting the question when it comes to permutations vs. combinations. While I understand the concept that permutations are for ordered arrangements, whereas combinations are not, I read this question as a permutation problem.

"How many different sets?" - makes me think that I need to solve for all potential arrangements that are unique since each candidate is inherently different and unlike the others.

Does anyone else have this problem w.r.t. interpretation????
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