Counting ways to distribute identical objects among finite number of bins Given positive integers $w_1,w_2,\ldots,w_n$ with $m=\sum_{i\in[n]}w_i$. We have $n$ bins numbered $1, 2, \ldots, n$ with capacity $w_1,w_2,\ldots,w_n$. In how many ways could we place $\lceil\frac{m}{2}\rceil$ identical objects in these $n$ bins, such that no bin holds more than its capacity?
I wonder if there's research about this quantity, e.g. upper and lower bounds.
 A: In general we can represent the possible fillings of bin $k$ as a polynomial
$$x^1+x^2+x^3+\cdots +x^{w_k}$$
interpreted as "bin $k$ may have $1$ object or $2$ objects or $3$ objects ... or $w_k$ objects". 
Each bin can be represented this way, so if we wish to represent combinations of all possibilities of bin fillings then it seems natural to write the product of each of these $n$ polynomials (the multiplication represents the logical "and" operation whereas "$+$" represents "or")
$$\prod_{k=1}^{n}\left(\sum_{r=1}^{w_k}x^r\right)$$
in the expansion of such a polynomial the coefficient of the term $x^{\lceil m/2\rceil}$ will receive a contribution of $1$ for every possible combination of valid bin fillings of our $n$ bins with $\lceil m/2\rceil$ identical objects. 
Hence in general the answer is
$$\left[x^{\lceil m/2\rceil}\right]\prod_{k=1}^{n}\left(\sum_{r=1}^{w_k}x^r\right)$$
where $\left[x^{\lceil m/2\rceil}\right]$ is the operator that extracts the  $x^{\lceil m/2\rceil}$ term of the expansion.
