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Counting Problem - Company recruitment

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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

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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

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Brent

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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

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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

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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.

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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

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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.

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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.
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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...

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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
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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