0

In how many ways can $n$ identical balls be distributed amongst $m$ different boxes given that a box can have any number of balls(from $0$ to $n$)?

What I've tried is using multinomial theorem to find the answer but that becomes a tedious task, here's what I made up:

For a given box, the can possibly be $0,1,2,3...(n-1),n$ balls.

Writing it in the form: $(x^0 + x^1 + x^2 + x^3 ... + x^{\text n-1} + x^n)$ Since there a $m$ boxes which such possibilities so they can multiplied and written as: $(x^0 + x^1 + x^2 + x^3 ... + x^{\text n-1} + x^n)^m$ And since in total there are $n$ balls, therefore we can find the coefficient of $x^n$ in the given product.

But that becomes a lengthy process for bigger values of $m$ and $n$, and I'm unable to generalize it.

Community
  • 1
Nikhil H
  • 1
  • Hi and welcome to the site! Since this is a site that encourages learning, you will get much more help if you show us what you have already done. Could you edit your question with your thoughts and ideas? – 5xum Sep 17 '14 at 07:38
  • How many ways can n balls be put into 1 box? 2 boxes? 3 boxes. Working through this should put you in good stead. – Nic Sep 17 '14 at 07:43
  • This link will be helpful as it contains answer:

    http://math.stackexchange.com/questions/618491/distribute-n-identical-objects-into-r-distinct-groups

     – spectraa Sep 17 '14 at 07:46
  • @WantTobeAbstract thanks that helped. – Nikhil H Sep 17 '14 at 07:56
  • @Nikhil H Take care, that the solution is in fact $\binom{n+k-1}{n}$ not $\binom{n+k-1}{n-1}$ possible ways as your problem differs slightly, in that the balls are identical, and therfore can be considered indistinguishable. Enumeration – Nic Sep 17 '14 at 08:17
  • @Nic aren't the balls identical in both cases. – spectraa Sep 17 '14 at 08:31
  • 1
    @WantTobeAbstract I agree it is confusing. Consider $\binom{4+2-1}{4}$ That is number of ways to put 4 balls into 2 boxes, which should be $5: (0,4), (1,3), (2,2), (3,1), (4,0)$. – Nic Sep 17 '14 at 08:57

0 Answers0