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Different ways in which donuts can be distributed

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by shrey2287 » Wed Sep 21, 2011 7:11 pm
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
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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.
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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.
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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.
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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

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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
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