Combination Problem(plz help)

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akpareek: Senior | Next Rank: 100 Posts; Posts: 38; Joined: Sun Feb 14, 2010 8:53 am

by akpareek » Fri Jul 30, 2010 5:09 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

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Source: Beat The GMAT — Problem Solving | Fri Jul 30, 2010 5:09 am

nithi_mystics: Senior | Next Rank: 100 Posts; Posts: 55; Joined: Mon Sep 07, 2009 9:11 pm; Location: California; Thanked: 4 times

by nithi_mystics » Fri Jul 30, 2010 5:18 am

akpareek wrote: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

If no one is eligible, 0 candidates will be chosen.. am i missing something here?

If all are eligible, then the required number of sets = 7C1 * 10C2 = 315

Thanks
Nithi

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selango: Legendary Member; Posts: 1460; Joined: Tue Dec 29, 2009 1:28 amFollowed by: 7 members; Thanked: 135 times

by selango » Fri Jul 30, 2010 5:46 am

7C1 * 10C2 = 315

--Anand--

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Gurpinder: Legendary Member; Posts: 659; Joined: Mon Dec 14, 2009 8:12 amFollowed by: 3 members; Thanked: 32 times

by Gurpinder » Fri Jul 30, 2010 7:46 am

akpareek wrote: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 best way to do a problem like this one that involves selecting people from 2 different groups is to calculate each one seperately and then multiply the two results you get.

so.....since this problem involves selecting people from a group, the order is not important and hence this is a combination problem.

so group 1: 7! / 1!6! = 7
group 2: 10!/2!8! = 45

7x45 = 315 and thats your answer.

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