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## 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?

- Thread starter jimmianlin
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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?

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lanedance

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have a look at

http://en.wikipedia.org/wiki/Combination

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- #4

lanedance

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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

- #5

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You mixed up the things my friend.

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

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.

[tex]P_{(3,2)} 5 = \frac{5!}{3!2!}=10[/tex]

For the second option.

1st 2nd 3rd 4th 5th

2 1 0 0 0

[tex]P_{(1,1,3)} 5 = \frac{5!}{3!1!1!}=20[/tex]

For the third option.

1st 2nd 3rd 4th 5th

3 0 0 0 0

[tex]P_{(1,4)} 5 = \frac{5!}{1!4!}=5[/tex]

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.

- #6

lanedance

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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, recieving 2 cookies is clearly different from recieving 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...

- #7

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You are right. Sorry for misunderstanding you.

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, recieving 2 cookies is clearly different from recieving 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...

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