# Thread: Probability of each combination with repetition

1. ## Probability of each combination with repetition

Hi guys,

It looks like trivial problem but I cannot solve it.

I have M baskets and N balls. I can choose one of the baskets randomly and put a ball into the chosen basket.

I'm just interested in combination. So in three baskets with 5 balls,

(1) (1) (3)

(1) (3) (1)

(3) (1) (1)

above combinations are same for me.

I just want to know the probability of each combinations. I know that all possible combination can be represented by combination with repetition.

But I do not know that how many cases of each combinations are existed.

Any helps are appreciated!!!

2. Originally Posted by ychang
I have M baskets and N balls. I can choose one of the baskets randomly and put a ball into the chosen basket.
(1) (1) (3); (1) (3) (1); (3) (1) (1)
(the)above combinations are same for me.
This a poorly worded question. Read it again considering my highlights.
Can a basket be empty? If not then there are just two possible answers for you example:
5=3+1+1 or 5=2+2+1.

If baskets can be empty you have: 5=5; =4+1; 5=3+1+1; or 5=2+2+1

So the answer depends upon what you mean.
Please respond with a clarification.

3. Thanks Plato,

Yes, basket can be empty. And I know that $\displaystyle \frac{(M+N-1)!}{(M-1)!N!}@$ ways are possible. Can I know that how many times each possible combinations are occured?

Thanks again!

4. Originally Posted by ychang
Yes, basket can be empty.
Then I think that you asking about the number of partitions of an integer in k or fewer summands.
Example: Say that there are seven balls and four baskets, then the answer is eleven.
7; 6+1, 5+2, 4+3; 5+1+1, 4+2+1, 3+2+2, 3+3+1; 4+1+1+1, 3+2+1+1, 2+2+2+1.

The formula is not easy. In fact it is given as a recursive function.