BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Different ways in which donuts can be distributed

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pm
Location: Milpitas, CA
Thanked: 1854 times
Followed by:523 members
GMAT Score:770

by Anurag@Gurome » Wed Sep 21, 2011 8:01 pm
shrey2287 wrote:Larry, Michael, and Doug have five donuts to share. If any one of the men can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed?
(A) 21
(B) 42
(C) 120
(D) 504
(E) 5040
This is a partition problem.
Consider the arrangement of two 1's and five zeros(1100000).
Here the zeros symbolize doughnuts.
Let us describe the arrangement as: Larry gets the zeros or doughnuts to the left of the first 1, Michael gets the zeros between the two 1's and Doug gets the zeros after the second 1.
So, 0000011 means that Larry gets the five doughnuts and Michael and Doug get none.
Similarly, 0100100 means that Larry gets 1 doughnut, Michael gets 2 and Doug gets 2.
So the total number ways of arranging 2 ones and 5 zeros will give us the required answer.
This is 7!/(2!*5!) = 21

The correct answer is A.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Thu Sep 22, 2011 4:06 am
shrey2287 wrote:Larry, Michael, and Doug have five donuts to share. If any one of the men can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed?
(A) 21
(B) 42
(C) 120
(D) 504
(E) 5040
I posted a solution here:

https://www.beatthegmat.com/experts-any- ... 82307.html

Before reading my explanation for this particular problem -- which appears at the bottom of the thread -- please read through my posts regarding two other problems solved with the same technique.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Senior | Next Rank: 100 Posts
Posts: 38
Joined: Tue Sep 13, 2011 6:48 am

by indi » Thu Sep 22, 2011 4:15 am
Hi,

Not able to understand. Can you please explain it again. I am struck at the 2 1's and 5 0's.

Thanks
Anurag@Gurome wrote:
shrey2287 wrote:Larry, Michael, and Doug have five donuts to share. If any one of the men can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed?
(A) 21
(B) 42
(C) 120
(D) 504
(E) 5040
This is a partition problem.
Consider the arrangement of two 1's and five zeros(1100000).
Here the zeros symbolize doughnuts.
Let us describe the arrangement as: Larry gets the zeros or doughnuts to the left of the first 1, Michael gets the zeros between the two 1's and Doug gets the zeros after the second 1.
So, 0000011 means that Larry gets the five doughnuts and Michael and Doug get none.
Similarly, 0100100 means that Larry gets 1 doughnut, Michael gets 2 and Doug gets 2.
So the total number ways of arranging 2 ones and 5 zeros will give us the required answer.
This is 7!/(2!*5!) = 21

The correct answer is A.
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 2623
Joined: Mon Jun 02, 2008 3:17 am
Location: Montreal
Thanked: 1090 times
Followed by:355 members
GMAT Score:780

by Ian Stewart » Thu Sep 22, 2011 6:23 am
I discussed a related problem here:

https://www.beatthegmat.com/combination-t41362.html

That said, I have *never* seen a 'partition problem' on the actual GMAT, or in any published source of official questions. While I think there is some non-zero chance the GMAT could ask such a question, it's extremely unlikely. Unless you have nothing else to study, I'd suggest you not spend time on this.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 158
Joined: Wed Jun 16, 2010 8:21 am
Thanked: 13 times
Followed by:1 members

by rohit_gmat » Thu Sep 22, 2011 7:35 am
shrey2287 wrote:Larry, Michael, and Doug have five donuts to share. If any one of the men can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed?
(A) 21
(B) 42
(C) 120
(D) 504
(E) 5040
I just love GMATGuruNY's bucket method, and once I read this Q, I knew that this was exactly a "bucket question".... so i read the problem in this way.. that 5 donuts go up in the air and the 3 guys open put up their buckets - and its possible that one guy gets all of them, just like the question says - and we need to find out in how many ways we can distribute the donuts..

so we have 5 donuts and 3 buckets (so thats 2 separators) ... OOOOO | | ... and we need to sort re-arrange this... (e.g. OO | O | OO .. 2 guys get 2 and 1 guy gets 1)....

so it is 7!/(5!2!) = 21


A
Top
Post new topic Post Reply
• Page 1 of 1