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# A certain company has 18 equally qualified applicants for 4

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Manager
Joined: 21 Mar 2007
Posts: 72
A certain company has 18 equally qualified applicants for 4 [#permalink]

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Updated on: 03 May 2014, 04:40
4
10
00:00

Difficulty:

5% (low)

Question Stats:

88% (00:47) correct 12% (00:52) wrong based on 475 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, 04:12.
Last edited by Bunuel on 03 May 2014, 04:40, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
Manager
Joined: 31 Oct 2007
Posts: 112
Location: Frankfurt, Germany

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07 Mar 2008, 04:18
1
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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Manager
Joined: 30 May 2010
Posts: 185
A certain company has 18 equally qualified applicants for 4 [#permalink]

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07 Sep 2010, 10:47
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
Math Expert
Joined: 02 Sep 2009
Posts: 46307
Re: GMATPrep 18 equally qualified applicants for 4 open position [#permalink]

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07 Sep 2010, 10:59
2
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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SVP
Joined: 08 Jul 2010
Posts: 2115
Location: India
GMAT: INSIGHT
WE: Education (Education)
A certain company has 18 equally qualified applicants for 4 [#permalink]

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09 Aug 2015, 06:19
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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Manager
Joined: 23 Sep 2015
Posts: 87
Concentration: General Management, Finance
GMAT 1: 680 Q46 V38
GMAT 2: 690 Q47 V38
GPA: 3.5
Re: A certain company has 18 equally qualified applicants for 4 [#permalink]

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22 Oct 2015, 13:51
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?
Current Student
Joined: 20 Mar 2014
Posts: 2643
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
A certain company has 18 equally qualified applicants for 4 [#permalink]

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22 Oct 2015, 14:00
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!
Manager
Joined: 05 Jul 2015
Posts: 105
GMAT 1: 600 Q33 V40
GPA: 3.3
Re: A certain company has 18 equally qualified applicants for 4 [#permalink]

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25 Nov 2015, 22:08
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
Manager
Joined: 03 Jan 2017
Posts: 180
Re: A certain company has 18 equally qualified applicants for 4 [#permalink]

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28 Mar 2017, 14:02
this is an easy combinatorics question

Combination formula will make it: 4C18=3060
Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2570
A certain company has 18 equally qualified applicants for 4 [#permalink]

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17 Jun 2018, 19:53
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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A certain company has 18 equally qualified applicants for 4   [#permalink] 17 Jun 2018, 19:53
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