Put B balls in C containers. How many combinations have box(es) with exactly 2 balls?

Question

Assume that we have B balls (all the same) and C numbered containers (distinguishable). We want to calculate how many of the total combinations contain exactly 1 container with 2 balls, exactly 2 containers with 2 balls, etc.

The way I see it, first step is to calculate the different ways that we can distribute the B balls in C containers, assuming that each container can have any number of balls, including zero. According to the Stars and bars paradigm, assuming that I am not making a mistake, this should be: $\dbinom{B+C-1}{C-1}$

Second step is to calculate how many of those combinations have:

• exactly 1 container with 2 balls
• exactly 2 containers with 2 balls
• exactly 3 containers with 2 balls
• etc

which I have no idea how to do and any leads or suggestions are greatly appreciated!

Answer 1

Hint: Try to use the generalized inclusion exclusion principle. http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle#A_generalization

Answer 2

Assume that there have to be exactly $c$ containers containing $2$ balls. Choose these $c$ containers in one of ${C\choose c}$ ways, and put two balls in each of them. The numbers $$C':=C-c, \quad B':=B-2c$$ remain fixed in the sequel.

There will be $m$ containers with $\geq3$ balls, where $m$ has to satisfy $$0\leq m\leq M:=\min\left\{\biggl\lfloor{B'\over3}\biggr\rfloor, C'\right\}\ .$$ Choose these $m$ containers in one of ${C'\choose m}$ ways, and put three balls in each of them. Now there are $C'-m\geq0$ containers still empty, and there are $B'-3m\geq0$ balls left over.

We can decide to put a single ball into $k$ of the empty containers, where $k$ has to satisfy $$0\leq k\leq k_m:=\min\{C'-m,B'-3m\}\ .$$ This can be done in ${C'-m\choose k}$ ways.

Now there are $B'-3m-k\geq0$ balls left, which can be freely distributed in the $m$ containers already holding $3$ balls. When $m\geq1$ this can be done in $${B'-3m-k+(m-1)\choose m-1}={B'-2m-k-1\choose m-1}$$ ways; the case $m=0$ is special, see below.

Putting it all together the requested number $N$ comes to $$N={C\choose c}\cdot\left({C'\choose B'}+\sum_{m=1}^M {C'\choose m}\sum_{k=0}^{k_m} {C'-m\choose k}{B'-2m-k-1\choose m-1}\right)\ .$$ I don't think that the double sum can be simplified.