We haveNbottles,mred andnblue balls,n > N,m > N.

We need to know in how many ways is it possible to partition

all read and blue balls into bottles. There should be at list

one red and one blue ball in each bottle.

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- May 9th 2012, 01:21 PMRafaelcombinatorics question
We have

**N**bottles,**m**red and**n**blue balls,**n > N**,**m > N**.

We need to know in how many ways is it possible to partition

all read and blue balls into bottles. There should be at list

one red and one blue ball in each bottle. - May 9th 2012, 01:28 PMPlatoRe: combinatorics question
- May 10th 2012, 11:09 AMRafaelRe: combinatorics question
- May 10th 2012, 12:40 PMPlatoRe: combinatorics question
There is an important concept known a

**integer partitions**

Because "There should be at least one red and one blue ball in each bottle".

We are partitioning the numbers $\displaystyle n-N~\&~m-N$

From that webpage, you can see that the is no simple formula or way of doing this problem.

Here is an example. Suppose $\displaystyle n=15,~m=10,~\&~N=6$.

We think this way. Go ahead and put a red in each bottle leaving 9 reds.

Go ahead and put a blue in each bottle leaving 4 blues.

Now there are $\displaystyle P(9,6)=26$ ways to partition 9 into 6 or fewer summands.

And $\displaystyle P(4,4)=5$ ways to partition 4 into 4 or fewer summands.

The product $\displaystyle 26\cdot 5=130$ is the number of ways of doing both. - May 10th 2012, 01:20 PMRafaelRe: combinatorics question
Thanks a lot