No. of positive integral solutions and link to coefficients in expansion

The formula they use is the same as Theorem One from Stars and Bars.

This is proved by considering $k$ fences that lie strictly within the fields, and only one fence between each pair of fields.

The 'zero-option' allows for fences to be placed outside the fields, and also more than one together, creating 'empty spaces'.

The generating function (GF) used is:

$$(x^1 + x^2 + \ldots )^r$$

For the zero-option, we would use:

$$(1 + x + x^2 + \ldots )^r$$

and for example if there were to be at least two fields between fences, we would use:

$$(x^2 +x^3 + \ldots )^r$$

These all go into the binomial expansion, and we look for the coefficients of $x^n, x^{n-r}, x^{n-2r}, \dots$, depending on the minimum value of $x_i$ allowed.

Which is like giving the minimum values to each player, do a zero-based stars and bars, and then give this as extra to each player.