Probability and combinations- Donuts

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venmic: Master | Next Rank: 500 Posts; Posts: 222; Joined: Mon Oct 13, 2008 4:04 pmFollowed by: 2 members; Thanked: 3 times

by venmic » Tue May 10, 2011 8:27 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

21

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Source: Beat The GMAT — Problem Solving | Tue May 10, 2011 8:27 pm

manpsingh87: Master | Next Rank: 500 Posts; Posts: 436; Joined: Tue Feb 08, 2011 3:07 amFollowed by: 6 members; Thanked: 72 times

by manpsingh87 » Tue May 10, 2011 10:32 pm

venmic 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

21

we have to distribute 5 donuts among 3 friends, it can be carried out in following ways;
case 1) 050; one of them receives 5 and other receives none, so total no. of ways of distributing the 5 donuts= 3!/2! (we have divided here by 2! because 2 people are getting same quantity);

case 2) 140; in this case total no. of distribution would be 3!=6; (because all are getting different donuts)
case 3) 230; in this case as well total no. of distribution would be 3!=6;
case 4) 221; in this case total no. of distribution would be 3!/2!=3;
case 5) 113; in this case total no. of distribution would be 3!/2!=3;

hence total required no. of distributions are= 3+6+6+3+3=21; hence A

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Tue May 10, 2011 11:46 pm

venmic 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

21

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.

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Thrills4ever: Junior | Next Rank: 30 Posts; Posts: 10; Joined: Thu May 12, 2011 12:47 pm; Thanked: 1 times

by Thrills4ever » Thu May 12, 2011 12:55 pm

Another way to do this is using the same separator method Mitch discusses in a previous post.

Let the arrangement of Donuts be written as D:

Since there are 5 Donuts, we will write DDDDD

Since there are 3 people, there can only be 2 separators, and we can show one like this: |

Either one person can get all 5 donuts or 0 or any combination.

Thus we can write D|D|DDD, we cannot write D|D|D|DD because there are only 3 people and if we wrote it like that, we would be adding an additional person (hopefully that makes sense)

The total number of separators is || or 2
The total number of donuts is DDDDD or 5

Thus we get 5+2 = 7..to reach the answer we simply write 7!/5!2! = 21

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smackmartine: Legendary Member; Posts: 516; Joined: Fri Jul 31, 2009 3:22 pmFollowed by: 13 members; Thanked: 112 times

by smackmartine » Thu May 12, 2011 6:41 pm

Can you please post the link where Mitch discussed separator method ? I remember that he did but could not find the same :(

Thrills4ever wrote:Another way to do this is using the same separator method Mitch discusses in a previous post.

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smackmartine: Legendary Member; Posts: 516; Joined: Fri Jul 31, 2009 3:22 pmFollowed by: 13 members; Thanked: 112 times

by smackmartine » Thu May 12, 2011 6:55 pm

Got it!!

https://www.beatthegmat.com/experts-any- ... tml#363310

smackmartine wrote:Can you please post the link where Mitch discussed separator method ? I remember that he did but could not find the same :(

Thrills4ever wrote:Another way to do this is using the same separator method Mitch discusses in a previous post.

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