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# A certain company has 18 equally qualified applicants for 4

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Manager
Joined: 21 Mar 2007
Posts: 56
A certain company has 18 equally qualified applicants for 4  [#permalink]

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Updated on: 03 May 2014, 03:40
4
22
00:00

Difficulty:

5% (low)

Question Stats:

87% (01:26) correct 13% (01:45) wrong based on 719 sessions

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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

Originally posted by japped187 on 07 Mar 2008, 03:12.
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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Joined: 02 Sep 2009
Posts: 64951
Re: GMATPrep 18 equally qualified applicants for 4 open position  [#permalink]

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07 Sep 2010, 09:59
3
2
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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##### General Discussion
Manager
Joined: 31 Oct 2007
Posts: 90
Location: Frankfurt, Germany

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07 Mar 2008, 03:18
2
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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09 Aug 2015, 05:19
1
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
1
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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GPA: 3.3
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
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Joined: 03 Jan 2017
Posts: 131
Re: A certain company has 18 equally qualified applicants for 4  [#permalink]

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28 Mar 2017, 13:02
this is an easy combinatorics question

Combination formula will make it: 4C18=3060
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A certain company has 18 equally qualified applicants for 4  [#permalink]

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17 Jun 2018, 18:53
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

Since order does not matter, 4 people can be chosen from 18 in:

18C4 = 18!/(4! x 14!) = (18 x 17 x 16 x 15)/4! = (18 x 17 x 16 x 15)/(4 x 3 x 2) = 3 x 17 x 4 x 15 = 3,060 ways.

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

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28 Jan 2019, 11:33
Hi All,

We're told that a certain company has 18 equally qualified applicants for 4 open positions. We're asked for the number of different groups of 4 applicants that can be chosen by the company to fill the positions if the order of selection does not matter. This question is a fairly straight-forward Combination Formula question (as a number of the other posts in this thread have shown). Once you recognize that that formula can be applied, you can actually avoid doing some of that math though:

The numerator of that Combination Formula calculation will include the number 17. Since that number is a prime number, there's nothing in the denominator that can 'reduce it' - meaning that the correct answer MUST be a multiple of 17. As such, you can quickly eliminate the first 3 answers (since they are clearly NOT multiples of 17). With a little work, you can also eliminate Answer D (since it's not a multiple of 17 either). That just leaves the correct answer...

GMAT assassins aren't born, they're made,
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Re: A certain company has 18 equally qualified applicants for 4   [#permalink] 28 Jan 2019, 11:33