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Probability and combinations- Donuts

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Probability and combinations- Donuts

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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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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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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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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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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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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