# How many combination can we create from this structure?

I 'm stuck in to this problem and I really need your help. Sorry if the title is not very informative. It is really hard for me to explain it in one sentence. I do my best in explaining it:

There exists these two natural integers:

m $\in \mathbb{N}$
$a_i \in \mathbb{N}$ for $i=1 \cdots m$

and some object with which I create a structure like this:
$$\matrix{ 1_1 & 1_2 & ... & 1_m \\ 1_1,2_1&1_2,2_2&...&1_m,2_m \\ 1_1,2_1,3_1 &1_2,2_2,3_2 & \ldots & 1_m,2_m,3_m \\ \vdots & \vdots & \vdots & \vdots \\ 1_1,2_1,...,a_1 & 1_2,2_2,...,a_2, &\ldots & 1_m,2_m,...,a_m\\ }$$

We can create new object by placing objects of different columns beside each other. All possible combinations has a general format like this:

$$(1_1),(1_2),...,(1_m)\\ (1_1),(1_2,2_2),...,(1_m)\\ ...\\ (1_1),(1_2),...,(1_m,2_m)\\ (1_1,2_1),(1_2),...,(1_m)\\ (1_1,2_1),(1_2,2_2),...,(1_m)\\ ...\\ (1_1,2_1),(1_2),...,(1_m,2_m)\\ ...\\ ...\\ (1_1,2_1,...,a_1),(1_2,2_2,...,a_2),...,(1_m,2_m,...,a_m)$$

I added the parenthesizes to make it more clear. Order is not important

In other words, I create all the combination of the objects in the first, second and ... and $a_1-th$ rows on the first column with the the first, second and ... and $a_2-th$ rows of the second column and so on. I each of these combinations there is an element from each and every column.

I think all possible combinations is $a_1 \times a_2 \times ... \times a_m$.

But, There is a restriction on number of objects in each of these combinations. let it be $$h\in \mathbb{N}$$ Number of objects in a combination can not be greater that $h$.

Now, the question is:
how many combinations do exist that satisfy this condition? (Order is not important)

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The problem is equivalent to the following. Let $N_k=\{1,\dots,a_k\}$ for $k=1,\dots,m$. Let $$G=\left\{\langle i_1,\dots,i_m\rangle\in\prod_{k=1}^mN_k:\sum_{k=1}^mi_k\le h\right\}\;;$$ you want to know $|G|$ in terms of $h,m$, and the numbers $a_k$ for $k=1,\dots,m$.

This is equivalent to asking for the number of solutions to the equation

$$x_1+\ldots+x_m+x_{m+1}=h-m\tag{1}$$

in non-negative integers subject to the upper bounds $x_k\le a_k-1$ for $k=1,\dots,m$: a solution $\langle x_1,\dots,x_{m+1}\rangle$ corresponds to $\langle x_1+1,\dots,x_m+1\rangle\in G$, with $x_{m+1}=h-\sum_{k=1}^mi_i$. Without the upper bounds this is a standard stars-and-bars problem, and $(1)$ has $\binom{h}m$ solutions in non-negative integers.

To account for the upper bounds, an inclusion-exclusion calculation is necessary. Replacing $x_k$ by $y_k=x_k-a_k$ and counting solutions in non-negative integers to

$$x_1+\ldots+x_{k-1}+y_k+y_{k+1}+\ldots+x_{m+1}=h-m-a_k$$

gives the number of solutions to $(1)$ that exceed the upper bound on $x_k$; by the stars-and-bars calculation that number is $\binom{h-a_k}m$. Subtracting these from $\binom{h}m$ gives the approximation $$\binom{h}m-\sum_{k=1}^m\binom{h-a_k}m\;.\tag{2}$$

This still isn’t correct, however, since it’s possible that some solutions to $(1)$ exceed more than one upper bound, and any such solution has been subtracted twice in $(2)$. If $1\le i<k\le m$, the number of solutions to $(1)$ that exceed the upper bounds on both $x_i$ and $x_k$ is $\binom{h-a_i-a_k}m$, and each of these numbers must be added back in to $(2)$ to yield the next approximation, $$\binom{h}m-\sum_{k=1}^m\binom{h-a_k}m+\sum_{1\le i<k\le m}\binom{h-a_i-a_k}m\;.$$ Continue in this fashion, alternately adding and subtracting the corrections. If we let $[m]^k$ denote the collection of $k$-element subsets of $\{1,\dots,m\}$, the final result is

$$\sum_{k=0}^m(-1)^k\sum_{S\in[m]^k}\binom{h-\sum\limits_{i\in S}a_i}m\;.\tag{3}$$

That’s a bit ugly, but if $h$ isn’t too large in comparison with the bounds $a_i$, most of the terms of $(3)$ will be $0$, since the upper number in the binomial coefficients will be non-positive for all but small values of $k$.

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