BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Counting Problem - Company recruitment

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 358
Joined: Thu Apr 18, 2013 9:46 am
Location: Jeddah, Saudi Arabia
Thanked: 42 times
Followed by:7 members
GMAT Score:730

Counting Problem - Company recruitment

by faraz_jeddah » Sat May 18, 2013 2:07 pm
A company organized a recruiting process for 3 vacant positions of assistant manager for its product launches. The positions are available in 3 different locations: Los Angeles, New York, and Boston. The company's efforts yielded 9 eligible candidates who were open to moving to any of the 3 available cities. How many ways can the company hire 3 out of the 9 eligible candidates for assistant manager positions in its 3 locations, assuming only one assistant manager per location?

A - 180
B - 325
C - 504
D - 730
E - 1260

OA is c
Top
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Sat May 18, 2013 2:40 pm
faraz_jeddah wrote:A company organized a recruiting process for 3 vacant positions of assistant manager for its product launches. The positions are available in 3 different locations: Los Angeles, New York, and Boston. The company's efforts yielded 9 eligible candidates who were open to moving to any of the 3 available cities. How many ways can the company hire 3 out of the 9 eligible candidates for assistant manager positions in its 3 locations, assuming only one assistant manager per location?

A - 180
B - 325
C - 504
D - 730
E - 1260

OA is c
Take the task of filling the 3 positions and break it into stages.

Stage 1: Select someone for the LA position
There are 9 candidates, so we can complete stage 1 in 9 ways.

Stage 2: Select someone for the New York position
There are 8 candidates remaining, so we can complete stage 2 in 8 ways.

Stage 3: Select someone for the Boston position
There are 7 candidates remaining, so we can complete stage 3 in 7 ways.

By the Fundamental Counting Principle (FCP) we can complete all 3 stages (and thus fill the 3 positions) in (9)(8)(7) ways ([spoiler]= 504 ways[/spoiler])

Answer: C

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 358
Joined: Thu Apr 18, 2013 9:46 am
Location: Jeddah, Saudi Arabia
Thanked: 42 times
Followed by:7 members
GMAT Score:730

by faraz_jeddah » Sun May 19, 2013 4:45 am
Thanks Brent!

Just realized I was making a multiplication error. :shock:
I was getting worked up because the problem looked straight forward. :D
Top
Junior | Next Rank: 30 Posts
Posts: 12
Joined: Thu Apr 11, 2013 11:02 pm
Thanked: 1 times

by rac.nishu » Fri Jun 07, 2013 12:33 am
Brent@GMATPrepNow wrote:
faraz_jeddah wrote:A company organized a recruiting process for 3 vacant positions of assistant manager for its product launches. The positions are available in 3 different locations: Los Angeles, New York, and Boston. The company's efforts yielded 9 eligible candidates who were open to moving to any of the 3 available cities. How many ways can the company hire 3 out of the 9 eligible candidates for assistant manager positions in its 3 locations, assuming only one assistant manager per location?

A - 180
B - 325
C - 504
D - 730
E - 1260

OA is c
Take the task of filling the 3 positions and break it into stages.

Stage 1: Select someone for the LA position
There are 9 candidates, so we can complete stage 1 in 9 ways.

Stage 2: Select someone for the New York position
There are 8 candidates remaining, so we can complete stage 2 in 8 ways.

Stage 3: Select someone for the Boston position
There are 7 candidates remaining, so we can complete stage 3 in 7 ways.

By the Fundamental Counting Principle (FCP) we can complete all 3 stages (and thus fill the 3 positions) in (9)(8)(7) ways ([spoiler]= 504 ways[/spoiler])

Answer: C

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775


I knw something is wrong with my approach but I wanna understand y?

Select 3 guys out of 9 -- 9C3 ..
And then send them to any of 3 locations . 3C1

so answer is 9C3* 3C1 = 252.. Why is this wrong approach? Pls explain .. will be helpful
Top
Senior | Next Rank: 100 Posts
Posts: 38
Joined: Fri Jan 04, 2013 12:55 am
Thanked: 11 times

by mkdureja » Fri Jun 07, 2013 1:54 am
I knw something is wrong with my approach but I wanna understand y?

Select 3 guys out of 9 -- 9C3 ..
And then send them to any of 3 locations . 3C1

so answer is 9C3* 3C1 = 252.. Why is this wrong approach? Pls explain .. will be helpful
Sending them to any of the three locations wont be 3C1.
Sending A, B, C to locations P, Q, R.
We aren't choosing 1 out of 3, but choosing 1 out of 3 and then choosing another 1 out of remaining 2., that'll be 3C1*2C1 = 6.
Top
Newbie | Next Rank: 10 Posts
Posts: 5
Joined: Mon Nov 05, 2012 2:11 am

by rkiran9 » Fri Jun 07, 2013 3:06 am
rac.nishu wrote:
Brent@GMATPrepNow wrote:
faraz_jeddah wrote:A company organized a recruiting process for 3 vacant positions of assistant manager for its product launches. The positions are available in 3 different locations: Los Angeles, New York, and Boston. The company's efforts yielded 9 eligible candidates who were open to moving to any of the 3 available cities. How many ways can the company hire 3 out of the 9 eligible candidates for assistant manager positions in its 3 locations, assuming only one assistant manager per location?

A - 180
B - 325
C - 504
D - 730
E - 1260

OA is c
Take the task of filling the 3 positions and break it into stages.

Stage 1: Select someone for the LA position
There are 9 candidates, so we can complete stage 1 in 9 ways.

Stage 2: Select someone for the New York position
There are 8 candidates remaining, so we can complete stage 2 in 8 ways.

Stage 3: Select someone for the Boston position
There are 7 candidates remaining, so we can complete stage 3 in 7 ways.

By the Fundamental Counting Principle (FCP) we can complete all 3 stages (and thus fill the 3 positions) in (9)(8)(7) ways ([spoiler]= 504 ways[/spoiler])

Answer: C

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775


I knw something is wrong with my approach but I wanna understand y?

Select 3 guys out of 9 -- 9C3 ..
And then send them to any of 3 locations . 3C1

so answer is 9C3* 3C1 = 252.. Why is this wrong approach? Pls explain .. will be helpful


Hi Brent,
I followed grid approach, i.e out of 9 people 3 needs to be selected means 6 are not selected..

so 9!/3!6!.... but its wrong..:(..

Regards
Ravikiran D
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 149
Joined: Wed May 01, 2013 10:37 pm
Thanked: 54 times
Followed by:9 members

by Atekihcan » Fri Jun 07, 2013 3:19 am
rkiran9 wrote:...I followed grid approach, i.e out of 9 people 3 needs to be selected means 6 are not selected..

so 9!/3!6!.... but its wrong..:(..
It is not just selecting 3 people out of 9 but after selecting you have to send the selected people in 3 different cities. So, you have to consider the possible number of arrangements among them which is 3!

So, the correct answer should be 9!/3!6! x 3! = 9!/6!

Hope that helps.
Top
Legendary Member
Posts: 512
Joined: Mon Jun 18, 2012 11:31 pm
Thanked: 42 times
Followed by:20 members

by sana.noor » Sat Jul 20, 2013 12:05 pm
just forget about three chosen people, remember their are 9 eligibale people and 3 vacant positions.
first seat at one location can be filled in 9 ways
second seat at second location in 8 ways and
3rd seat at third location in 7 ways
so the answer will be 9.8.7 = 504 ways

However if the question asks to choose three members out of 9 eligible candidates and order isnt important than the answer will be 9.8.7/1.2.3 = 84 different combinations of 3 members.
Work hard in Silence, Let Success make the noise.

If you found my Post really helpful, then don't forget to click the Thank/follow me button. :)
Top
Newbie | Next Rank: 10 Posts
Posts: 2
Joined: Tue Jul 23, 2013 10:32 pm

by GmatPrepK » Wed Jul 24, 2013 11:52 pm
Initially lets keep aside the location requirement and just figure out the number of 3 member combinations for 3 vacant posts from the 9 candidates available.

it comes to 9C3 = 9!/(9-3)! * 3! = 84. So we have 84 possible 3 candidate combinations.

The 3 candidates (Say A, B and C) for each combination can be placed at 3 locations in the following 6 ways:


Los Angeles || New York || Boston.

1 . A || B || C
2 . A || C || B
3 . B || A || C
4 . B || C || A
5 . C || A || B
6 . C || B || A

i.e 3 candidates can be places in 3 places in 3P3 = 3! /(3-3!) = 6 ways.


Hence the total number of ways the company can select 3 candidates for 3 locations is 84*6 = 504 ways.


Hope this helps...
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 164
Joined: Sat Sep 20, 2014 10:26 pm
Thanked: 1 times

by jaspreetsra » Sun Dec 28, 2014 1:29 am
A company organized a recruiting process for 3 vacant positions of assistant manager for its product launches. The positions are available in 3 different locations: Los Angeles, New York, and Boston. The company's efforts yielded 9 eligible candidates who were open to moving to any of the 3 available cities. How many ways can the company hire 3 out of the 9 eligible candidates for assistant manager positions in its 3 locations, assuming only one assistant manager per location?

A - 180
B - 325
C - 504
D - 730
E - 1260
LA: 9 ways
NY: 8 ways
B: 7 ways
9*8*7 = 504
Hard work brings success!
Top
User avatar
Newbie | Next Rank: 10 Posts
Posts: 6
Joined: Wed Jul 15, 2015 5:39 am

by Johnniewales » Sat Dec 05, 2015 3:07 pm
Ravikiran the issue about your approach is that you missed the second step, The first step of 9C3 is correct but the next step was to use the FCP for unique
Objects which we had three options in step 2 I.e 3x2x1.
So we have step one multiplied by step 2.
9C3 x 3x2x1= 84x6=504
Top
Post new topic Post Reply
• Page 1 of 1