In how many ways can we placen+xballs innboxes

with a condition thatat least 1 ball is present in every box

when

1)balls r identical

2)balls r numbered from 1 to n+x

and x<n

Explain ur answer

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- Dec 14th 2011, 05:47 AMlivinggourmandnumber of ways ?
In how many ways can we place

**n+x**balls in**n**boxes

with a condition that__at least 1 ball is present in every box__

when

1)balls r identical

2)balls r numbered from 1 to n+x

and x<n

Explain ur answer - Dec 14th 2011, 07:10 AMPlatoRe: number of ways ????
You failed to tell us if the boxes are all different.

So we assume they are.

There are $\displaystyle \binom{K+N-1}{K}$ ways to place K identical items into N different cell.

If we require that no cell is empty then it must be that case that $\displaystyle K\ge N$ and that multi-selection formula becomes

$\displaystyle \binom{K-1}{K-N}$.

For part b), we need to count the number of surjections (onto functions) from a set of N+x to a set of N.

If $\displaystyle K\ge N$ and then the number of surjections from a set of K to a set of N is $\displaystyle \sum\limits_{j = 0}^N {( - 1)^j \binom{N}{j} (N - j)^K } $. - Dec 14th 2011, 07:19 AMlivinggourmandRe: number of ways ????
n if boxes are identical??

- Dec 14th 2011, 07:39 AMPlatoRe: number of ways ????
If the boxes are identical then it becomes more difficult.

For part a), we have to count the numbers of ways to have K summands for the integer N+x. That is not easy.

Part b) is a bit easier that a). We count the number of unordered partitions of N+n individuals into N groupings.

However, I suspect that whoever wrote this question meant the boxes all different. Because, that is an easier question.