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Permutaion and combination

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by josh80 » Wed Dec 11, 2013 5:29 pm
A certain company has 18 equally qualified applicants for 4 open positions. How many different groups of 4 applicants can be chosen by company to fill the positions if the order of selection doesn't matter?

18
72
180
1260
3060
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by Brent@GMATPrepNow » Wed Dec 11, 2013 5:47 pm
josh80 wrote:A certain company has 18 equally qualified applicants for 4 open positions. How many different groups of 4 applicants can be chosen by company to fill the positions if the order of selection doesn't matter?

A) 18
B) 72
C) 180
D) 1260
E) 3060
Since the order of the selected applicants does not matter, we can use combinations to solve this question.
We can select 4 applicants from 18 applicants in 18C4 ways.

We have a free video on calculating combinations in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789
In this video, we learn how to first take a combination like 18C4 and rewrite it as ...
(18)(17)(16)(15)/(4)(3)(2)(1) We now need to EVALUATE this.
First recognize that we can simplify some parts to get: (18)(17)(2)(5)

IMPORTANT: Since the answer choices are so SPREAD APART, we can likely use some estimation, rather than evaluate (18)(17)(2)(5)

First notice that, (18)(17)(2)(5) = (18)(170)
Now notice that (10)(170) = 1700, which means (18)(170) must be GREATER than 1700
So, the correct answer must be greater than 1700

Choose E

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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