# combinatorics question

• May 9th 2012, 02:21 PM
Rafael
combinatorics 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, 02:28 PM
Plato
Re: combinatorics question
Quote:

Originally Posted by Rafael
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.

Two essential questions:
1) are the bottles identical or distinct?
2) are the balls identical( except for color) or distinct?
• May 10th 2012, 12:09 PM
Rafael
Re: combinatorics question
Quote:

Originally Posted by Plato
Two essential questions:
1) are the bottles identical or distinct?
2) are the balls identical( except for color) or distinct?

All bottles are identical, all red balls are identical, all blue balls are identical.
• May 10th 2012, 01:40 PM
Plato
Re: combinatorics question
Quote:

Originally Posted by Rafael
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.

Quote:

Originally Posted by Rafael
All bottles are identical, all red balls are identical, all blue balls are identical.

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 $n-N~\&~m-N$
From that webpage, you can see that the is no simple formula or way of doing this problem.
Here is an example. Suppose $n=15,~m=10,~\&~N=6$.
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 $P(9,6)=26$ ways to partition 9 into 6 or fewer summands.
And $P(4,4)=5$ ways to partition 4 into 4 or fewer summands.
The product $26\cdot 5=130$ is the number of ways of doing both.
• May 10th 2012, 02:20 PM
Rafael
Re: combinatorics question
Thanks a lot