| View previous topic :: View next topic |
| Author |
Message |
gibran Rising GMAT Star
Joined: 12 Apr 2008 Posts: 42
Thanks given: 0 Thanked 1 times in 1 posts
Test Date: 10 June 2009 Target GMAT Score: 770
|
Posted: Thu May 15, 2008 7:28 am Post subject: Assign employees |
|
|
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 |
|
| Back to top |
|
 |
|
|
punit.kaur.mba Really wants to Beat The GMAT!
Joined: 13 Feb 2008 Posts: 104
Thanks given: 1 Thanked 7 times in 7 posts
Test Date: Not Decided Yet Target GMAT Score: 750+
|
Posted: Thu May 15, 2008 7:41 am Post subject: |
|
|
Either the 2question is incorrectly framed and should also consider the possibility that none of the employess are assigned to any office. That way the answer would be 5.
Else the answer should be 4 and i dont see that in the options!!
the 2 offices
Off 1, Off 2
No. of ways-
Off 1 - 3 , Off2 0
off1 - 2 , Off2 - 1
Off1 - 0, Off2 -3
Off1 - 1, Off2 - 2
Someone correct me if I m wrong! _________________ http://punitkaur.blogspot.com/ |
|
| Back to top |
|
 |
aatech Really wants to Beat The GMAT!
Joined: 29 Apr 2008 Posts: 191
Thanks given: 0 Thanked 11 times in 10 posts
|
Posted: Thu May 15, 2008 8:11 am Post subject: |
|
|
IMO D
Case 1 - One office has all 3 employees and second has none - this can happen in 2 ways
Case 2 - One office has 1 employee and another office has 2 - this can happen in 6 ways
So total 8 |
|
| Back to top |
|
 |
Magellan Rising GMAT Star
Joined: 10 Mar 2008 Posts: 59
Thanks given: 3 Thanked 2 times in 2 posts
Test Date: TBD Target GMAT Score: 730+
|
Posted: Thu May 15, 2008 8:27 am Post subject: |
|
|
| aatech wrote: | | Case 2 - One office has 1 employee and another office has 2 - this can happen in 6 ways |
How is that possible? I see only two cases: 1 - 2 or 2 - 1 |
|
| Back to top |
|
 |
netigen GMAT Destroyer!
Joined: 18 Feb 2008 Posts: 633
Thanks given: 3 Thanked 23 times in 23 posts
|
Posted: Thu May 15, 2008 11:38 am Post subject: |
|
|
0 | 3 -> 1 way to do this
1 | 2 -> 3 ways to do this
2 | 1 -> 3 ways to do this
3 | 0 -> 1 way to do this
-------------
Total = 8 (assuming each employee is a unique individual 8)) |
|
| Back to top |
|
 |
|
|
punit.kaur.mba Really wants to Beat The GMAT!
Joined: 13 Feb 2008 Posts: 104
Thanks given: 1 Thanked 7 times in 7 posts
Test Date: Not Decided Yet Target GMAT Score: 750+
|
Posted: Thu May 15, 2008 11:54 am Post subject: |
|
|
Actually aatech is right. The answer is 8.
We are right in selecting cases but wrong in counting the number of arrangements in that case.
In second case
(2,1) and (1,2)
Each case inturn has 3 ways becos we are "arranging" people between 2 offices here.
So if E denotes Employee and O denotes office
Selecting E1 and E2 is not same as selecting E2 and E3.
So within the second case
1) 3 ways of arranging people in O1,O2
(a) O1 - E1 and O2-E2,E3
(b)O1-E2 and O2-E1,E3
(c)O1-E3 and O2-E1,E2
2) similar to above case but just reversed
(a) O1 - E2,E3, O2 - E1
(b) O1- E1,E3 O2-E2
(c)O1-E1,E2 O2- E3 _________________ http://punitkaur.blogspot.com/ |
|
| Back to top |
|
 |
Magellan Rising GMAT Star
Joined: 10 Mar 2008 Posts: 59
Thanks given: 3 Thanked 2 times in 2 posts
Test Date: TBD Target GMAT Score: 730+
|
Posted: Thu May 15, 2008 12:11 pm Post subject: |
|
|
| OK, I see. Difference between permutation and combination if I am not mistaken. |
|
| Back to top |
|
 |
ektamatta Really wants to Beat The GMAT!
Joined: 13 May 2008 Posts: 151
Thanks given: 2 Thanked 1 times in 1 posts
Location: dallas,tx usa Target GMAT Score: 650
|
Posted: Thu May 15, 2008 6:33 pm Post subject: |
|
|
there is another way to solve this problem
2 office
3 employees
2^3
8 ways |
|
| Back to top |
|
 |
lunarpower GMAT Instructor

Joined: 03 Mar 2008 Posts: 530
Thanks given: 0 Thanked 69 times in 66 posts
GMAT Score: 800
|
Posted: Fri May 16, 2008 10:38 pm Post subject: |
|
|
| ektamatta wrote: | there is another way to solve this problem
2 office
3 employees
2^3
8 ways |
that works, although it's certainly not self-explanatory enough as is.
here's a fuller explanation:
let's call the offices 1 and 2.
for each employee, there's a choice of being assigned either to office 1 or to office 2. furthermore, wherever the previous employees are assigned has no effect upon the assignments of the other employees, since double-/triple-packing and empty offices are both fine.
therefore:
3 employees
2 options for each employee
multiply the #s of options, because they're independent sequential choices: 2 x 2 x 2 = 8 total sets of choices. _________________ ron purewal
instructor, mgmat
Learn more about me |
|
| Back to top |
|
 |
umaa Really wants to Beat The GMAT!
Joined: 08 Jun 2008 Posts: 117
Thanks given: 2 Thanked 0 times in 0 posts
|
Posted: Wed Jun 25, 2008 4:27 am Post subject: easy way |
|
|
| An easy way to find it. You have an equation to find this if there are no conditions. That is, n^r. which is 2^3=8. |
|
| Back to top |
|
 |
|
|
|