Is this the right mindset for this problem?

Expert replies

kellogs4toniee: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Thu Jan 05, 2012 7:21 am

by kellogs4toniee » Sat Jan 21, 2012 11:54 am

Q: 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

The book gave an explanation that I only understood about 80%. This is how I solved the question: There are two positions. First position has 7C1 combinations (7). Second position has 10C2 combinations (45). Then you multiply 7 and 45 to get 315 total combinations of both positions.

That got me the right answer. Is that the right way and mindset I should be doing these type of problems? Is there something that I missed?

Thanks,

Tony

Source: Beat The GMAT — Problem Solving | Sat Jan 21, 2012 11:54 am

LalaB: Master | Next Rank: 500 Posts; Posts: 425; Joined: Wed Dec 08, 2010 9:00 amFollowed by: 7 members; Thanked: 56 times; GMAT Score:690

by LalaB » Sat Jan 21, 2012 12:12 pm

I did it in the same way as u did.
my question is what that book offers to do? please show us how it is solved in the book.

kellogs4toniee: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Thu Jan 05, 2012 7:21 am

by kellogs4toniee » Sat Jan 21, 2012 12:21 pm

LalaB wrote:I did it in the same way as u did.
my question is what that book offers to do? please show us how it is solved in the book.

The part in the book that I did not fully understand says this:

To fill the positions in the computer science department, any one of the 10 eligible candidates can be chosen for the first position and any of the remaining 9 can be chosen for the second, making 10x9 = 90 sets of 2 candidates to fill the computer science position. But, this number includes the set in which candidate A was chosen to fill the first position and B was chosen to fill the second as well as the set in which B was chosen for the first, and A was chosen for the second. Therefore, there are 90/2 = 45 sets of 2 candidates.

After re-reading it, I think I understand now. The bolded part is basically defining the reason why we are using combinations instead of permutations. The order does not matter in the question, and the book tried explaining that through the words and terms of the problem.

LalaB: Master | Next Rank: 500 Posts; Posts: 425; Joined: Wed Dec 08, 2010 9:00 amFollowed by: 7 members; Thanked: 56 times; GMAT Score:690

by LalaB » Sat Jan 21, 2012 12:29 pm

yep, partially the same logic is used. u can write 10C2 (10!/2!*8!=10*9/2). or 10*9/2 (the method used in the book) . everything is clear now

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