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# A certain company assigns employees to offices in such a way

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Senior Manager
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A certain company assigns employees to offices in such a way [#permalink]

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26 Sep 2007, 12:53
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A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 3 employees to 2 different offices?

A. 5
B. 6
C. 7
D. 8
E. 9

I suppose this is the easiest problem one can expect on GMAT, still i got it wrong..........i picked B, please help! What concept i'm missing over here?
CEO
Joined: 29 Mar 2007
Posts: 2583
Followers: 19

Kudos [?]: 420 [0], given: 0

Re: DS - SET 23 - Q6 [#permalink]

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26 Sep 2007, 13:05
singh_amit19 wrote:
A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 3 employees to 2 different offices?

A. 5
B. 6
C. 7
D. 8
E. 9

I suppose this is the easiest problem one can expect on GMAT, still i got it wrong..........i picked B, please help! What concept i'm missing over here?

I don' think this is easy at all. Ive never seen a combinatorics Q like this before.

Is the answer E? Im stuck btwn D or E, but id vouch for E. I can't really explain it, its just what I got in my notes.

we have 3!/2! * 3!/2! = 9. But i really don't know if this is correct.
Senior Manager
Joined: 11 Sep 2005
Posts: 324
Followers: 2

Kudos [?]: 238 [0], given: 0

Re: DS - SET 23 - Q6 [#permalink]

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26 Sep 2007, 13:08
GMATBLACKBELT wrote:
singh_amit19 wrote:
A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 3 employees to 2 different offices?

A. 5
B. 6
C. 7
D. 8
E. 9

I suppose this is the easiest problem one can expect on GMAT, still i got it wrong..........i picked B, please help! What concept i'm missing over here?

I don' think this is easy at all. Ive never seen a combinatorics Q like this before.

Is the answer E? Im stuck btwn D or E, but id vouch for E. I can't really explain it, its just what I got in my notes.

we have 3!/2! * 3!/2! = 9. But i really don't know if this is correct.

If i keep aside all my technical knowledge of P&C, i'll land up with E though the OA is not E
Director
Joined: 03 May 2007
Posts: 886
Schools: University of Chicago, Wharton School
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Kudos [?]: 190 [0], given: 7

Re: DS - SET 23 - Q6 [#permalink]

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26 Sep 2007, 13:23
singh_amit19 wrote:
A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 3 employees to 2 different offices?

A. 5
B. 6
C. 7
D. 8
E. 9

I suppose this is the easiest problem one can expect on GMAT, still i got it wrong..........i picked B, please help! What concept i'm missing over here?

D. 8

x: 0 - 1 - 2 - 3
y: 3 - 2 - 1 - 0

if x = 0, y = 3 => 1 way
if x = 1, y = 2 => 3 ways
if x = 2, y = 1 => 3 ways
if x = 3, y = 0 => 1 way

so total = 8 ways
CEO
Joined: 29 Mar 2007
Posts: 2583
Followers: 19

Kudos [?]: 420 [0], given: 0

Re: DS - SET 23 - Q6 [#permalink]

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26 Sep 2007, 13:52
Fistail wrote:
singh_amit19 wrote:
A certain company assigns employees to offices in such a way that some of the offices can be empty and more than one employee can be assigned to an office. In how many ways can the company assign 3 employees to 2 different offices?

A. 5
B. 6
C. 7
D. 8
E. 9

I suppose this is the easiest problem one can expect on GMAT, still i got it wrong..........i picked B, please help! What concept i'm missing over here?

D. 8

x: 0 - 1 - 2 - 3
y: 3 - 2 - 1 - 0

if x = 0, y = 3 => 1 way
if x = 1, y = 2 => 3 ways
if x = 2, y = 1 => 3 ways
if x = 3, y = 0 => 1 way

so total = 8 ways

Nice, i kinda had this, but I really didn't think it was right.
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5062
Location: Singapore
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Kudos [?]: 356 [0], given: 0

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26 Sep 2007, 18:34
Case 1: empty, all 3 in one office -> 2 ways
Case 2: 1 employee, 2 employees -> 2 * 3C2 = 6 ways
Total = 8 ways
26 Sep 2007, 18:34
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