Distributing $k$ distinct items among $r$ distinct groups without ordering 
Calculate the number of ways of distributing $k$ distinct items
  among $r$ distinct groups such that each group receives at least
  $a$ and at most $b$ items and internal arrangement of items
  within groups doesn't matter.

For example suppose there are 2 groups and 3 items A, B, C. The distributions (AB, C) and (BA, C) must not be counted twice.
I've read a similar question here. But the solution doesn't satisfy my second condition.
 A: Let $a,b,r$ be fixed.  For any $k$,  let $f(k)$ be the number of ways of arranging the $k$ distinct items into $r$ distinct groups with the specifications you give.  The exponential generating function for $f(k)$ is
$$
\sum_{k=0}^\infty f(k) \frac{x^k}{k!} = \left( \frac{x^a}{a!} + \frac{x^{a+1}}{(a+1)!} + \cdots + \frac{x^b}{b!} \right)^r
$$
So, for a fixed $k = k_0$, $f(k_0)$ is the coefficient of $\frac{x^k}{k!}$ in the power series above.  An explicit formula is
$$
f(k_0) = \frac{d^k}{dx^k} \left. \left[ \left( \frac{x^a}{a!} + \frac{x^{a+1}}{(a+1)!} + \cdots + \frac{x^b}{b!} \right)^r \right] \right|_{x = 0}
$$
I highly doubt there is any way to simplify this in general with so many parameters involved.  However, if you pose specific conditions on $a$ and $b$ you get some nicer results.  For example if $a = 0$ and $b = \infty$, then
the generating function becomes $(e^x)^r = e^{rx}$, so $f(k) = r^k$.
But you probably already knew that.
Addendum:
If we require that $|b - a| \le 1$, then either $b = a$ or $b = a+1$.


*

*If $b = a$, we want the coefficient of $\frac{x^k}{k!}$ in $\left(\frac{x^a}{a!}\right)^r$, which should be easy.

*If $b = a+1$, we want the coefficient of $\frac{x^k}{k!}$ in $\left(\frac{x^a}{a!} + \frac{x^{a+1}}{(a+1)!}\right)^r = \frac{x^{ar}}{(a!)^r} \left(1 + \frac{x}{a+1}\right)^r$.
This ultimately evaluates to
$$
{r \choose k - ar} \cdot \frac { k! } { (a!)^r (a+1)^{k - ar} }
$$
