distributing r identical balls into n distinct boxes w/ at l

I have a distribution problem.

How many ways can we distribute r identical objects into n distinct boxes with at least TWO in each box?.

I know the number of total ways is C(r+n-1,r) and with at least one in each box is C(r-1,n-1).

The opposite of at least two would be at least one and none, right?. But then what if, say, one box was empty, or two boxes empty and the rest distributed accordingly?.

Re: distirbuting balls among boxes.

I know that you realize that \(\displaystyle 2r \ge n\).
Go ahead and put two balls into each box.
Then distribute the remaining 2 r – n balls into the \(\displaystyle n\) boxes.

Re: distirbuting balls among boxes.

Hey pka. I reckon I may be over thinking this. I tend to do that with these sorts of problems.

Let's say I have 15 identical balls and 4 distinct boxes. I put two in each box. That leaves 7 to distribute in C(7+4-1,7)=C(10,7)=120 ways.

Is that all there is to it?. I got this answer with a generating function but did not trust it. I used \(\displaystyle \left(\sum_{k=2}^{9}x^{k}\right)^{4}\)

Then looked at the coefficient of the 15th power and it was 120. I used 9 because that is the most we can have in one box and still have at least two in the other 3 boxes.

Re: distirbuting balls among boxes.

Yes, that is it. It is quite simple.
You really should get Ivan Nivin's little book on counting.
It is an MAA publication.

Re: distirbuting balls among boxes.

Thank you. I will get the book. I like counting and prob problems. They can be challenging.

Re: distirbuting balls among boxes.

OK you 2 counters(!):
6 identical objects, 3 boxes.
The 6 objects are put in the boxes at random.
Probability that each box will contain 2?

Possibilities:
0,0,6
0,1,5
0,2,4
0,3,3
1,1,4
1,2,3
2,2,2

WHY isn't the probability 1/7 ?

The first 2 objects will end up in same box 1/3 of time, or in different boxes 2/3 of time, right?

Re: distirbuting balls among boxes.

Ok, ok...think I'm beginning to see "the light"; ran program (a few million tries) and probabilities are (roughly):
0,0,6 : .0042
0,1,5 : .0495
0,2,4 : .1235
0,3,3 : .0824
1,1,4 : .1235
1,2,3 : .4934 (probably 1/2)
2,2,2 : .1235
total : 1.0000

Sooooo? I'm now ashamed of jumping to conclusion that each group would be 1/7, and will repent by reciting
a few times: "Sir Skeeter, who art in heaven, ......"