Discrete Math: Distributing 11 Cookies to 50 Children - Efficient Solution?

[jimmianlin]

Homework Statement

If I want to know how many ways there are to distribute 11 chocolate chip cookies to 50 children, is there any way to do this without brute force?

Homework Equations

The Attempt at a Solution

 

[lanedance]

yes, basically you want to choose 11 children from 50 without any ordering

have a look at
http://en.wikipedia.org/wiki/Combination

 

[jimmianlin]

Its actually not that simple. Let me clarify the question. The chocolate chip cookies are all identical, but the students are distinct. For instance, one cookie could be given to each student, or all cookies could be given to one child.

 

[lanedance]

ok so you want combinations with reptition then, which is towards the bottom of the wiki page

this site has a reasonable explanation of how to derive the formula at teh very end of the page
http://www.mathsisfun.com/combinatorics/combinations-permutations.html

 

[njama]

ok so you want combinations with reptition then, which is towards the bottom of the wiki page

this site has a reasonable explanation of how to derive the formula at teh very end of the page
http://www.mathsisfun.com/combinatorics/combinations-permutations.html

You mixed up the things my friend.

He got 11 cookies and 50 children. He need to distribute the 11 cookies on different ways to 50 children.

I will give you example for smaller amounts.

Lets say you got 3 cookies and 5 children.

3 cookies can be distributed:

- 1 cookie for three persons 1*3=3

- 2 cookies for one person and 1 for one of the others 2+1=3

- 3 cookies for one person, 0 for the others 3+0=3

For the first option.

1st 2nd 3rd 4th 5th
1 1 1 0 0

Now make permutations with repetition, because IT DOES matter which children will get a cookie and which not.

P_{(3,2)} 5 = \frac{5!}{3!2!}=10

For the second option.

1st 2nd 3rd 4th 5th
2 1 0 0 0

P_{(1,1,3)} 5 = \frac{5!}{3!1!1!}=20

For the third option.

1st 2nd 3rd 4th 5th
3 0 0 0 0

P_{(1,4)} 5 = \frac{5!}{1!4!}=5

So the total number of distributions would be 10+20+5=35

I believe that my theory is correct. Please correct me if I am wrong. Thank you.

 

[lanedance]

Hi njama, not sure what you mean was mixed up?

When referring to ordering, I was implying the cookies are indistinguishable for all intents and purposes (ie. receieving a cookie is the same as receiving any other cookie). This makes it a combination rather than permutation question.

As you and jimmianlin point out, receiving 2 cookies is clearly different from receiving 1 cookie... You could treat this with a repetitive approach of working out the combinations of each cookie distribution, as you have done, but this will become difficult with large cookie/children numbers

That leads you to looking at combinations with repetition (see previous website for good explanation)

In your case
n = 5 number of children
r = 3 number of cookies
(n-1+r)!/((n-1)!r!) = 7!/(4!3!) = 7.5 = 35, agreeing with your work

I feel like a cookie after all this...

 

[njama]

Hi njama, not sure what you mean was mixed up?

When referring to ordering, I was implying the cookies are indistinguishable for all intents and purposes (ie. receieving a cookie is the same as receiving any other cookie). This makes it a combination rather than permutation question.

As you and jimmianlin point out, receiving 2 cookies is clearly different from receiving 1 cookie... You could treat this with a repetitive approach of working out the combinations of each cookie distribution, as you have done, but this will become difficult with large cookie/children numbers

That leads you to looking at combinations with repetition (see previous website for good explanation)

In your case
n = 5 number of children
r = 3 number of cookies
(n-1+r)!/((n-1)!r!) = 7!/(4!3!) = 7.5 = 35, agreeing with your work

I feel like a cookie after all this...

You are right. Sorry for misunderstanding you.