Ten identical cookies are to be distributed among five different kids (A, B, C, D, and E). All 10 cookies are distributed. How many different ways can the five kids be given cookies?

Question

reemii · Answer

but i want a cookie.

luigi0210 · Answer

I was never here

dan815 · Answer

look how we can seperate cookies with places like separating by sticks |dw:1370586947943:dw|

luigi0210 · Answer

I'll take reemii's cookie

dan815 · Answer

it shud let me post more questions than 1 at a time -.-

anonymous · Answer

I used a simplified version of some of the methods posted above. I thought about dividing up the 10 cookies as “making cuts.” For instance, you have 10 cookies, so there are 9 possible cuts that can be made in the spaces between those 10 cookies. If you make 0 cuts, then 1 person will be given all the cookies. If you make 1 cut, then 2 people will receive cookies. If you make 2 cuts, then 3 people will receive cookies…..If you make 4 cuts, then all 5 students will receive cookies. Or more generally, if you make n cuts, then n+1 people will receive cookies.

So the problem is just summing up all the ways that the cookies can be distributed when 0,1,2,3, and 4 cuts are made. This is a simple summation.

For 0 cuts, we want to know how many ways we can choose 0 cuts from 9 possible cuts and then how many ways we can distribute the resulting section of cookies to the 5 people. The product is the number of ways that all 10 cookies can be distributed to one of the people. So we calculate 9 CHOOSE 0 * 5 CHOOSE 1 and this will give us the total number of ways that all 10 cookies can be given to one person.

For 1 cut (remember that with 1 cut, 2 people will receive cookies), we calculate 9 CHOOSE 1 * 5 CHOOSE 2.

For 2 cuts (3 people receive cookies), we calculate 9 CHOOSE 2 * 5 CHOOSE 3.

You’ve probably already noticed a pattern. In order to distribute cookies to n people we need to make (n – 1) cuts. So to figure out the number of ways that cookies can be distributed to n people, we just calculate 9 CHOOSE (n – 1) * 5 CHOOSE n.

And now we are ready for our summation:

5
Σn = 1 ( 9 CHOOSE (n – 1) * 5 CHOOSE n )

Sorry about the weird looking notation.