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Probability and combinations

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by venmic » Sun Jul 03, 2011 12:15 am
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
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by Anurag@Gurome » Sun Jul 03, 2011 12:43 am
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?
We can use the 'stick separator' method here.

As there are three persons we have to divide the 5 donuts in three groups, where one group may not contain any donuts at all. Let us represent each donut by '*' and each separator by '|'. Hence some of the possible groupings are...
  • (0, 1, 4) --> |*|****
    (2, 1, 2) --> **|*|**
    (5, 0, 0) --> *****|| etc
Therefore number of such groupings = Number of ways to arrange 5 donuts and 2 separators = (5 + 2)!/[(5!)*(2!)] = 7!/[(5!)*(2!)] = 7*6/2 = 21
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by vishwas.arora » Mon Aug 01, 2011 6:46 am
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 ! :lol:

Thanks
Vishwas
Jai Hind !
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by Ozlemg » Mon Aug 01, 2011 2:07 pm
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
The more you suffer before the test, the less you will do so in the test! :)
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