A certain company has 18 equally qualified applicants for 4 : GMAT Problem Solving (PS)
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# A certain company has 18 equally qualified applicants for 4

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Manager
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A certain company has 18 equally qualified applicants for 4 [#permalink]

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07 Mar 2008, 03:12
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A certain company has 18 equally qualified applicants for 4 open positions. How many different groups of 4 applicants can be chosen by the company to fill the positions if the order of selection does not matter?

(A) 18
(B) 72
(C) 180
(D) 1,260
(E) 3,060
[Reveal] Spoiler: OA

Last edited by Bunuel on 03 May 2014, 03:40, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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07 Mar 2008, 03:18
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This is a combination question since order is not important.

n!
--------
k! (n-k!)

= 18!
--------
4! 14!

= 18 * 17 * 16 * 15 * 14!
---------------------------
4 * 3 * 2 * 1 * 14!

= 3060 (Ans: E)
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A certain company has 18 equally qualified applicants for 4 [#permalink]

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07 Sep 2010, 09:47
A certain company has 18 equally qualified applicants for 4 open positions. How many different groups of 4 applicants can be chosen by the company to fill the positions if the order of selection does not matter?

(A) 18
(B) 72
(C) 180
(D) 1,260
(E) 3,060
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Re: GMATPrep 18 equally qualified applicants for 4 open position [#permalink]

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07 Sep 2010, 09:59
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jpr200012 wrote:
A certain company has 18 equally qualified applicants for 4 open positions. How many different groups of 4 applicants can be chosen by the company to fill the positions if the order of selection does not matter?

(A) 18
(B) 72
(C) 180
(D) 1,260
(E) 3,060

# of ways to choose 4 different people out of 18, when order of chosen people doesn't matter is $$C^4_{18}=\frac{18!}{14!*4!}=3060$$.

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Re: A certain company has 18 equally qualified applicants for 4 [#permalink]

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02 May 2014, 10:44
Hello from the GMAT Club BumpBot!

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Re: A certain company has 18 equally qualified applicants for 4 [#permalink]

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06 Jun 2015, 23:31
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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A certain company has 18 equally qualified applicants for 4 [#permalink]

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09 Aug 2015, 05:19
japped187 wrote:
A certain company has 18 equally qualified applicants for 4 open positions. How many different groups of 4 applicants can be chosen by the company to fill the positions if the order of selection does not matter?

(A) 18
(B) 72
(C) 180
(D) 1,260
(E) 3,060

Initially There are 18 choices for first Position
Now, There are 17 choices for Second Position
Now, There are 16 choices for Third Position
Now, There are 15 choices for Forth Position

i.e. Total Ways to choose people (With arrangement) = 18*17*16*15

Since we require only the selection hence we need to exclude the arrangements of 4 selected individuals which is 4!

i.e. i.e. Total Ways to choose people (WithOUT arrangement) = (18*17*16*15)/4! = 3060

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Re: A certain company has 18 equally qualified applicants for 4 [#permalink]

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22 Oct 2015, 12:51
Bunuel wrote:
jpr200012 wrote:
A certain company has 18 equally qualified applicants for 4 open positions. How many different groups of 4 applicants can be chosen by the company to fill the positions if the order of selection does not matter?

(A) 18
(B) 72
(C) 180
(D) 1,260
(E) 3,060

# of ways to choose 4 different people out of 18, when order of chosen people doesn't matter is $$C^4_{18}=\frac{18!}{14!*4!}=3060$$.

if order mattered how would this problem change?
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A certain company has 18 equally qualified applicants for 4 [#permalink]

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22 Oct 2015, 13:00
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GMATDemiGod wrote:
Bunuel wrote:
jpr200012 wrote:
A certain company has 18 equally qualified applicants for 4 open positions. How many different groups of 4 applicants can be chosen by the company to fill the positions if the order of selection does not matter?

(A) 18
(B) 72
(C) 180
(D) 1,260
(E) 3,060

# of ways to choose 4 different people out of 18, when order of chosen people doesn't matter is $$C^4_{18}=\frac{18!}{14!*4!}=3060$$.

if order mattered how would this problem change?

If the order mattered, then the 4 people you selected out of 18 via 18C4 need to be multiplied by 4 to account for the fact that 4 positions will themselves be arranged in 4! ways.

Thus total ways possible with an ordered set = 18C4* 4!
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Re: A certain company has 18 equally qualified applicants for 4 [#permalink]

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25 Nov 2015, 21:08
18 * 17 * 16 * 15
Is a little bigger than 20*15*15*15

Divide away 4*3*2

5*5*15*15/2

Little more than close to 3,000

Tip: Multiplying by 15 is quick and easy. Add a 0 (*10) and then add half of the number
Re: A certain company has 18 equally qualified applicants for 4   [#permalink] 25 Nov 2015, 21:08
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