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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- faraz_jeddah
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- Brent@GMATPrepNow
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Take the task of filling the 3 positions and break it into stages.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

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

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Just realized I was making a multiplication error.

I was getting worked up because the problem looked straight forward.

Brent@GMATPrepNow wrote:Take the task of filling the 3 positions and break it into stages.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

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

Sending them to any of the three locations wont be 3C1.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 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.

rac.nishu wrote:Brent@GMATPrepNow wrote:Take the task of filling the 3 positions and break it into stages.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

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

- Atekihcan
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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!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....

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

Hope that helps.

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.

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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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...

- jaspreetsra
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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

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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- Johnniewales
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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