venmic wrote:Reopening thread...


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?


Please explain

Just an alternate explanation of how I approached the problem,

Let Larry, Michael and Doug get x, y and z doughnuts respectively,
clearly, x+y+z = 5

Provided, x varies from 0 to 5, y varies from 0 to 5 and z varies from 0 to 5 [each x, y and z taking only whole numbers 0,1,2,3,4 or 5]

Now we need to find the number of solutions (x,y,z) such that these conditions are satisfied

We know x+y+z=5 is an equation of a plane and we need to find the number of whole number points (x,y,z) on this plane lying in the first quadrant.

Let's break this plane into number of lines taking each cross-section,

For z=0, this becomes x+y=5 --> clearly giving 6 solutions (0,5), (1,4), (2,3), (3,2), (4,1) and (5,0)
For z=1, this becomes x+y=4 --> clearly giving 5 solutions (0,4), (1,3), (2,2), (3,1) and (4,0)

Similarly till z=5, we get total number of solutions 6+5+4+3+2+1 = 6(6+1)/2 = 21 whole number solutions of (x,y,z)

And this is what is required to be found !

Hope this helps !

Thanks
Vishwas

venmic wrote:Reopening thread...


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?


Please explain

Hi
There is a formula for this kind of problems.
The total number of ways dividing m identical object into R groups--> n+r-1 C r-1

n : Donuts: 5
r : People : 3

5+3-1 C 5-1 --> 7!/5!2! --> 21