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by phelps » Sat Jul 17, 2010 4:21 pm
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

OAe

please explain
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by debmalya_dutta » Sat Jul 17, 2010 4:54 pm
number of ways 1 candidate can be selected from 7 = 7
number of ways 2 candidate can be selected from 10 = 10C2
Total number of ways = 10C2* 7=315
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by Rahul@gurome » Sat Jul 17, 2010 5:10 pm
Number of ways of selecting 1 candidate from 7 candidates = 7C1 = 7
Number of ways of selecting 2 candidate from 10 candidates = 10C2 = 45
Therefore, required number of sets of 3 candidates = 7 * 45 = 315

The correct answer is (E).
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by Brent@GMATPrepNow » Thu Jan 09, 2020 6:06 am
phelps 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

OAe

please explain
We can take the task of filling both positions and break it into stages.

Stage 1: Fill the 1 math position
There are 7 candidates, so we can complete stage 1 in 7 ways

Stage 2: Fill the 2 computer science positions
Note that the order in which we select candidates doesn't matter. For example, selecting candidate B and then candidate C is the same as selecting candidate C and then candidate B. So, we can use combinations here.
We can select 2 candidates from 10 in 10C2 ways (= 45 ways)

Aside: If anyone is interested, here's a video on calculating combinations (like 10C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

By the Fundamental Counting Principle (FCP), we can complete the two stages (and thus fill the three positions in (7)(45) ways (= 315 ways)

Answer: E
--------------------------

Note: the FCP can be used to solve the majority of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
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- https://www.beatthegmat.com/simple-one-t270061.html
- https://www.beatthegmat.com/mouse-pellets-t274303.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
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- https://www.beatthegmat.com/digits-numbers-t270127.html
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DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
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- https://www.beatthegmat.com/permutation- ... 73915.html
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- https://www.beatthegmat.com/laniera-s-co ... 15764.html

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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