The total number of ways to put $N$ distinct balls into $K$ distinct boxes so that every box has more than one ball (assuming that $N \geqslant 2K$)

Question

So i came across the general combinatoric problem as stated in the title. I have $N$ distinct balls and $K$ distinct boxes $\left( N \geqslant 2K \right)$. I need to find the total number of ways to arrange these balls in these boxes so that every box has more than one ball.

Any help would be greatly appreciated. Thank you.

Answer 1

Here's a start with inclusion-exclusion.

Let $P_j^0$ be all ball-box arrangements such that the $jth$ box contains exactly $0$ balls, and let $P_j^1$ be similarly those in which the $jth$ box contains exactly 1 ball.

The total number of ball-box arrangements is $K^N$, so the number you are looking for is $K^N - \big| \bigcup\limits_{j=1}^{K} P_j^0 \cup P_j^1 \big|$. Inclusion exclusion is how you'll approach that big union.

You can attack it straight on because the number of ball-box arrangements having $J$ particular boxes empty and disjoint $I$ particular boxes containing exactly one ball is easily calculated. We pick $I$ of our $N$ balls and permute them into the designated boxes, and then accounting for the $J$ boxes that are off limits, we have $N-I$ balls to put into $K - J - I$ boxes. In other words, if $A_I$ and $A_J$ are disjoint subsets of $[1,\ldots,K]$ with $|A_I| = I$ and $|A_J| = J$, then

$$ \big| \bigcap\limits_{j \in A_J} P_j^0 \cap \bigcap\limits_{i \in A_I} P_i^1 \big| = \frac{N!}{(N-I)!} (K - I - J)^{N-I} $$

To account for all of the ways to choose $J$ and $I$ such boxes, you have a multiplicative factor of ${{K} \choose {I}} {{K-I} \choose {J}}$

The theorem of inclusion exclusion gives you that

$$ \big| \bigcup\limits_{j=1}^{K} P_j^0 \cup P_j^1 \big| = \sum_{A_I,A_J \subset [1,\ldots,K] \\ A_I \cap A_J = \emptyset} (-1)^{I+J+1} \big| \bigcap\limits_{j \in A_J} P_j^0 \cap \bigcap\limits_{i \in A_I} P_i^1 \big| $$

so with the above observations you should be all set! (we're summing over all ways to pick two disjoint subsets of the boxes.)

This post on Stirling numbers of the second kind may be helpful, since this application of inclusion-exclusion is just a slightly more complicated version of that one.

Answer 2

For Inclusion-Exclusion, let us count the number of ways that $j$ of the $K$ bins have less than $2$ of the $N$ balls. We will break things down into cases where $i$ of the bins have $1$ ball and $j-i$ have $0$ balls:

Choose the $j$ bins to have less than $2$: $\binom{K}{j}$
Choose $i$ of those $j$ bins to have $1$ ball: $\binom{j}{i}$
Choose $i$ of the $N$ balls for those $i$ bins: $\binom{N}{i}$
Choose an order for those $i$ balls: $i!$
Fill in the remaining $K-j$ bins with the remaining $N-i$ balls: $(K-j)^{N-i}$

Therefore, $$ N(j)=\sum_{i=0}^j\binom{K}{j}\binom{j}{i}\binom{N}{i}i!(K-j)^{N-i}\tag{1} $$ Thus, $$ \newcommand{\stirtwo}[2]{\left\{#1\atop#2\right\}} \begin{align} \sum_{j=0}^K(-1)^jN(j) &=\sum_{j=0}^K(-1)^j\sum_{i=0}^j\binom{K}{j}\binom{j}{i}\binom{N}{i}i!(K-j)^{N-i}\tag{2}\\ &=\sum_{j=0}^K(-1)^j\sum_{i=0}^j\binom{K}{i}\binom{K-i}{j-i}\binom{N}{i}i!\sum_m\stirtwo{N-i}{m}\binom{K-j}{m}m!\tag{3}\\ &=\sum_{i=0}^K\binom{K}{i}\binom{N}{i}i!\sum_m\stirtwo{N-i}{m}m!\sum_{j=i}^K(-1)^j\binom{K-i}{j-i}\binom{K-j}{m}\tag{4}\\ &=\sum_{i=0}^K\binom{K}{i}\binom{N}{i}i!\sum_m\stirtwo{N-i}{m}m!\sum_{j=i}^K(-1)^j\binom{K-i}{K-j}\binom{K-j}{m}\tag{5}\\ &=\sum_{i=0}^K\binom{K}{i}\binom{N}{i}i!\sum_m\stirtwo{N-i}{m}m!\sum_{j=i}^K(-1)^j\binom{K-i}{m}\binom{K-i-m}{K-j-m}\tag{6}\\ &=\sum_{i=0}^K\binom{K}{i}\binom{N}{i}i!\stirtwo{N-i}{K-i}(K-i)!(-1)^i\tag{7}\\ &=\bbox[5px,border:2px solid #C0A000]{K!\sum_{i=0}^K(-1)^i\binom{N}{i}\stirtwo{N-i}{K-i}}\tag{8} \end{align} $$ Explanation:
$(2)$: Inclusion-Exclusion and $(1)$
$(3)$: $\binom{K}{j}\binom{j}{i}=\binom{K}{i}\binom{K-i}{j-i}$ and $(K-j)^{N-i}=\sum\limits_m\stirtwo{N-i}{m}\binom{K-j}{m}m!$
$(4)$: rearrange terms and order of summation
$(5)$: $\binom{K-i}{j-i}=\binom{K-i}{K-j}$
$(6)$: $\binom{K-i}{K-j}\binom{K-j}{m}=\binom{K-i}{m}\binom{K-i-m}{K-j-m}$
$(7)$: $\sum\limits_{j=i}^K(-1)^j\binom{K-i-m}{K-j-m}=(-1)^i[m=K-i]$
$(8)$: $\binom{K}{i}i!(K-i)!=K!$

where the $\stirtwo{n}{k}$ are the Stirling Numbers of the Second Kind.

Answer 3

The combinatorial species here is the labeled species $\mathfrak{S}_{=K}(\mathfrak{P}_{\ge 2}(\mathcal{Z}))$ which gives the generating function

$$(\exp(z)-z-1)^K.$$

The desired statistic is then given by

$$N! [z^N] (\exp(z)-z-1)^K = N! [z^N] \sum_{q=0}^K {K\choose q} (-1)^q z^q (\exp(z)-1)^{K-q} \\ = N! \sum_{q=0}^K {K\choose q} (-1)^q [z^{N-q}] (\exp(z)-1)^{K-q} \\ = N! K! \sum_{q=0}^K \frac{1}{q!} (-1)^q [z^{N-q}] \frac{(\exp(z)-1)^{K-q}}{(K-q)!} \\= N! K! \sum_{q=0}^K \frac{1}{q! (N-q)!} (-1)^q (N-q)! [z^{N-q}] \frac{(\exp(z)-1)^{K-q}}{(K-q)!} \\= K! \sum_{q=0}^K {N\choose q} (-1)^q {N-q\brace K-q}.$$

This matches the answer by @robjohn.

Answer 4

First find the number of ways to put 2 balls into every box. Since the balls are distinguishable, there are $N\choose{2K}$ ways to pick $2K$ balls, and there are $\frac{(2K)!}{2^K}$ ways to divvy those $2K$ balls into the $K$ distinguishable boxes (think of lining up all $2K$ balls where the first two go into box 1, the next two into box 2, etc. There are $(2K)!$ possible line-ups, but the order of the $K$ pairs does not matter, so we should divide by $2^K$)

Then, for the remaining $N-2K$ balls you have $K^{N-2K}$ options to distribute those.

Now multiply all those together:

$$\frac{{N\choose{2K}}(2K)!K^{N-2K}}{2^K}$$

EDIT

This does NOT work, since it overcounts: I could initially pick a pair of balls A and B to go into some box, and later add ball C, but I could also initially pick pair A and C, and later add B. This ends up with the same balls in that box, but the above calculation counts these as different. :(

I'll leave this flawed formula up here until someone finds the actual correct formula ... Maybe it's still of some use, if only to show what does NOT work!