Probability and combinations- Donuts

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

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

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.

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.