|
Author |
Message |
|
TAGS:
|
|
|
Manager
Joined: 22 Jul 2009
Posts: 177
Location: Manchester UK
|
A certain company assigns employees to offices in such a way [#permalink]
Show Tags
 08 Jan 2010, 14:18
2
This post received
KUDOS
30
This post was
BOOKMARKED
D
E
Question Stats:
57% (00:59) correct 43% (01:07) wrong based on 589 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
|
|
|
|
|
|
Math Expert
Joined: 02 Sep 2009
Posts: 44600
|
Re: A certain company [#permalink]
Show Tags
08 Jan 2010, 16:58
5
This post received
KUDOS
Expert's post
2
This post was
BOOKMARKED
|
|
|
|
|
|
Manager
Joined: 22 Jul 2009
Posts: 177
Location: Manchester UK
|
Re: A certain company [#permalink]
Show Tags
09 Jan 2010, 03:27
i am still not able to understand. Can you please explain in detail?
also please tell me where i went wrong.This was my logic.
No. of people
office 1: 0|0|0|1|1|1|2|2|3
office 2: 1|2|3|0|1|2|0|1|0
this gives me 9 possible combination
|
|
|
|
|
|
Math Expert
Joined: 02 Sep 2009
Posts: 44600
|
Re: A certain company [#permalink]
Show Tags
09 Jan 2010, 04:25
3
This post received
KUDOS
Expert's post
6
This post was
BOOKMARKED
sagarsabnis wrote: i am still not able to understand. Can you please explain in detail?
also please tell me where i went wrong.This was my logic.
No. of people
office 1: 0|0|0|1|1|1|2|2|3
office 2: 1|2|3|0|1|2|0|1|0
this gives me 9 possible combination First of all you should assign ALL 3 employees to either of the offices. You can have the following scenarios: No. of people ***********A|B|C|D| office 1: 0|1|2|3| office 2: 3|2|1|0| In scenario (A) and (D) there is only one way to assign three people. But in (B) and (C) there will be 3 cases in each: Let's say there are 3 employees: Tom, Mary and Kate. In (B): Tom can be in office #1 and Mary/Kate in #2 OR Mary can be in #1 and Tom/Kate in #2 OR Kate in #1 and Tom/Mary in #2. Total 3 cases for (B). The same for (C). (A)+(B)+(C)+(D)=1+3+3+1=8. The way I solved this was different: Each of the three employees, Tom, Mary and Kate, has two choices office #1 or office #2. Hence total # of combinations (assignments) is 2*2*2=2^3=8. Hope it's clear.
_________________
New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!
Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.
Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics
|
|
|
|
|
|
Manager
Joined: 08 Sep 2010
Posts: 201
Location: India
WE 1: 6 Year, Telecom(GSM)
|
Re: A certain company employee [#permalink]
Show Tags
Updated on: 25 Oct 2010, 04:43
7
This post received
KUDOS
2
This post was
BOOKMARKED
Two offices can be filled in two ways, when all the three employee will be in same room or when two employee in one room and one in other room.
when all the three employee will be in same = 3C3 * 2! =2 (2!, because any of the room can be taken)
when two employee in one room and one in other room. = 3C2 * 1C1 * 2! = 6
Hence total ways = 6+2
Answer is 8.
Consider KUDOS if it is helpful to u.
Originally posted by ankitranjan on 25 Oct 2010, 04:26.
Last edited by ankitranjan on 25 Oct 2010, 04:43, edited 1 time in total.
|
|
|
|
|
|
Manager
Joined: 16 Oct 2010
Posts: 85
Location: United States
Concentration: Finance, Entrepreneurship
WE: Information Technology (Investment Banking)
|
Re: A certain company employee [#permalink]
Show Tags
25 Oct 2010, 04:38
monirjewel 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 Every employee has got the possibilit of getting assigned to any of the two offices. Hence total possibilities = 2^3 = 8
|
|
|
|
|
|
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8028
Location: Pune, India
|
Re: Why D any not B? please help me out [#permalink]
Show Tags
17 Nov 2010, 08:56
1
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
SoniaSaini 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
thanks in advance!!! For each one of the 3 employees, there are two choices. He can be allotted to any one of the two offices. Hence total number ways will be 2 * 2* 2 = 8 ways
_________________
Karishma
Veritas Prep | GMAT Instructor
My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews
|
|
|
|
|
|
Senior Manager
Joined: 29 Sep 2009
Posts: 386
|
Re: A certain company [#permalink]
Show Tags
21 Nov 2010, 07:12
4
This post received
KUDOS
4
This post was
BOOKMARKED
The best way to remember this is :
(Decisions) ^ (Players)
For this problem - 2 decisions , 3 players : 2^3=8
|
|
|
|
|
|
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8028
Location: Pune, India
|
Re: Emplyoees [#permalink]
Show Tags
28 Nov 2011, 02:43
1
This post received
KUDOS
Expert's post
2
This post was
BOOKMARKED
ashiima wrote: Hi,
I am kind of lost on all probability type qs :/
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 Think in this way: There is no restriction on the offices i.e. they can be vacant, they can accommodate all 3 employees etc. But there is a restriction on the employees i.e. each one of them must get an office. Employee 1 can get an office in 2 ways - office A or office B Employee 2 can get an office in 2 ways - office A or office B Employee 3 can get an office in 2 ways - office A or office B All three can be allotted offices in 2*2*2 = 8 ways This takes care of all cases.
_________________
Karishma
Veritas Prep | GMAT Instructor
My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews
|
|
|
|
|
|
Manager
Joined: 25 May 2011
Posts: 136
|
Re: Emplyoees [#permalink]
Show Tags
13 Dec 2011, 01:29
Attachment: 
gmat2.jpg [ 66.67 KiB | Viewed 4150 times ]
|
|
|
|
|
|
Magoosh GMAT Instructor
Joined: 28 Nov 2011
Posts: 302
|
Re: Emplyoees [#permalink]
Show Tags
13 Dec 2011, 11:12
The fastest way to solve this problem is by using the formula, 2^n, where n stands for the number of elements, or, in this case, the number of employees. This formula is derived from adding the number of combinations from What’s important with this problem is not to treat it as a probability problem. While on the surface it may seem similar to a typical combinations problem, using the combinations formula to solve the problem is cumbersome. Instead, use the formula, 2^n, where n stands for the number of elements, or, in this case, the number of employees. This formula is derived from adding the number of combinations whenever you can select any number greater than zero and less than or equal to n. For instance, here we could have chosen any of three employees for the first office. So instead of using 3C0 + 3C1 + 3C2 + 3C3, we can use 2^3. This formula becomes especially useful for larger numbers. Imagine the question were: How many ways can 8 employees go in two offices?
(A) 8
(B) 32
(C) 48
(D) 64
(E) 120Following the method of finding each case would take too much time. By using 2^n, we 2^8 = 64. (D) Going back to my original point: do not think of this as a typical probability problem, but one that uses the 2^n concept. The problems, while not really a probability problem, employ the 2^n formula. Positive integer N is the product of three distinct primes. How many factors in N?Ans: 2^3 – 1 (zero is not a factor so hence we subtract one from 2^n). A multiple-choice test has five possible answer choices. Any number of answers can be correct. (e.g. A-B-D is possible answer, C-D, or all five). How many different possible answers?Ans: 2^5 – 1 = 31 You can’t leave question blank (like the empty office) so therefore -1. By understanding the concept behind a question, instead of grouping a question under one general category, you should be able to solve problems more quickly.
_________________
Christopher Lele
Magoosh Test Prep
|
|
|
|
|
|
Manager
Joined: 02 Apr 2012
Posts: 76
Location: United States (VA)
Concentration: Entrepreneurship, Finance
WE: Consulting (Consulting)
|
Re: A certain company assigns employees to offices in such a way [#permalink]
Show Tags
06 Jul 2013, 17:59
1
This post received
KUDOS
Thank you. Then, if the company assigns employees to offices in such a way that if the offices can not be empty and more than one employee can be assigned to an office. And we have 5 employees and 3 rooms, the answer would be: 120? I mean, 5!
_________________
Encourage cooperation! If this post was very useful, kudos are welcome 
"It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James
|
|
|
|
|
|
Math Expert
Joined: 02 Sep 2009
Posts: 44600
|
Re: A certain company assigns employees to offices in such a way [#permalink]
Show Tags
07 Jul 2013, 01:13
2
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
|
|
|
|
|
|
Manager
Joined: 02 Apr 2012
Posts: 76
Location: United States (VA)
Concentration: Entrepreneurship, Finance
WE: Consulting (Consulting)
|
Re: A certain company assigns employees to offices in such a way [#permalink]
Show Tags
07 Jul 2013, 13:35
Thank you Bunuel! I have difficulties learning combinations, this is my weakest area in the GMAT. I am planning to practice all the combinations problems in the forum. Regarding the problem that I have posted before, I think that you mean: \(3^5\) - the combinations in which zero is an element in the set and it cannot be zero in any of the slots, with the restrictions that the 3 elements must sum up 5): {(005),(014),(023),(032),(041) ; (050)(140),(230),(320),(410) ; (500),(104),(203),(302),(401)} 243 - 15 = 228 I tried to apply combinatorics formulas to this problem (because writing that set is very time consuming) but I could not figure it out. Translating the problem: I need to find the number of combinations of three digits in which at least one of the digits is "0", the sum of those three digits is 5 and the digits range from 0 to 5 (six elements). Then, substract this number from \(3^5\)
_________________
Encourage cooperation! If this post was very useful, kudos are welcome
"It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James
|
|
|
|
|
|
Manager
Joined: 02 Apr 2012
Posts: 76
Location: United States (VA)
Concentration: Entrepreneurship, Finance
WE: Consulting (Consulting)
|
Re: A certain company assigns employees to offices in such a way [#permalink]
Show Tags
10 Jul 2013, 12:40
Applying combinations I think would be in this way: \(C^4_1 * C^2_1 = 4*2 = 8\)
_________________
Encourage cooperation! If this post was very useful, kudos are welcome
"It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James
|
|
|
|
|
|
Manager
Joined: 04 Dec 2011
Posts: 70
|
Re: A certain company assigns employees to offices in such a way [#permalink]
Show Tags
31 Aug 2013, 07:47
ok, can someone tell me what's wrong with my thinking.. 1st office can have any 3 employees.. therefore 3 options, 2nd office can also have any of 3 employees hence again 3 options so it should be 3*3=9 i think the logic is similar to the way Bunuel did..the only difference is in that case we had 2 choices for each employee therefore it was 2*2*2=8.. but why is the answer different in both cases?
_________________
Life is very similar to a boxing ring.
Defeat is not final when you fall down…
It is final when you refuse to get up and fight back!
1 Kudos = 1 thanks
Nikhil
|
|
|
|
|
|
Math Expert
Joined: 02 Sep 2009
Posts: 44600
|
Re: A certain company assigns employees to offices in such a way [#permalink]
Show Tags
03 Sep 2013, 06:42
|
|
|
|
|
|
Manager
Joined: 10 Apr 2016
Posts: 55
Concentration: Strategy, Entrepreneurship
GPA: 3.01
|
Re: A certain company assigns employees to offices in such a way [#permalink]
Show Tags
18 Apr 2016, 12:28
1
This post was
BOOKMARKED
Three people that can go in either office 1 or 2 So 2^3 =8 Important is to see that you are not distributing offices to people. That would be 3^2 instead of 2^3
_________________
Took the Gmat and got a 520 after studying for 3 weeks with a fulltime job. Now taking it again, but with 6 weeks of prep time and a part time job. Studying every day is key, try to do at least 5 exercises a day.
|
|
|
|
|
|
Director
Joined: 07 Dec 2014
Posts: 962
|
A certain company assigns employees to offices in such a way [#permalink]
Show Tags
18 Apr 2016, 15:46
with 2 offices and 3 employees, there are 3 ways to have one's own office, 3 ways to share an office with one other, and 2 ways to share an office with two others, or 8 ways total
Attachments
office.jpg [ 34.22 KiB | Viewed 8477 times ]
|
|
|
|
|
|
Intern
Joined: 08 Sep 2016
Posts: 4
|
Re: A certain company assigns employees to offices in such a way [#permalink]
Show Tags
05 Nov 2016, 06:26
Hi
I solved it the following way. Is my approach correct? Please comment.
4*1*2=8
we have 4 options ( 0,1,2,3 employees) for the first office and 1 for the other. Multiply by 2 as we can assign employees to 2nd office first ,in which case 1st office will have only 1 option.
B
|
|
|
|
|
|
|
Re: A certain company assigns employees to offices in such a way
[#permalink]
05 Nov 2016, 06:26
|
|
|
|
|
Go to page
1 2
Next
[ 28 posts ]
|
|
|
|
|
|
close
