The number of ways in which $n$ distinct items can be divided among $r$ groups 
The number of ways in which $n$ distinct items can be divided among $r$ groups such that no  group contains less than $m$ and not more than $k$ items $(m<k)$ is

Please solve this question.I am having no idea how to solve this question because I hadn't started Binomial Chapter ,I was wondering if there is any way to solve this question using Permutation and Combination.
Edit:-I tried but to solve this question but I couldn't reach anywhere.Please answer this question and try to understand  I am not trying to be lazy to ask homework question.
 A: Duplicating  the   construction  for  Stirling  numbers   we  get  the
combinatorial class
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SET}_{=r}(\textsc{SET}_{m\le\cdot\le k}
(\mathcal{Z})).$$
The corresponding EGF is 
$$\frac{1}{r!} \left(\sum_{q=m}^k \frac{z^q}{q!} \right)^r.$$
The desired quantity is then given by
$$a_{n,r} = n! [z^n]
\frac{1}{r!} \left(\sum_{q=m}^k \frac{z^q}{q!} \right)^r.$$
If a recurrence is wanted with memoization we write
$$a_{n,r} = \frac{1}{r} n! [z^n]
\sum_{p=m}^k \frac{z^p}{p!} \frac{1}{(r-1)!}
\left(\sum_{q=m}^k \frac{z^q}{q!} \right)^{r-1}
\\ = \frac{1}{r} n!
\sum_{p=m}^k \frac{1}{p!} [z^{n-p}] \frac{1}{(r-1)!}
\left(\sum_{q=m}^k \frac{z^q}{q!} \right)^{r-1}
\\ = \frac{1}{r} n!
\sum_{p=m}^k \frac{1}{p!} \frac{1}{(n-p)!} a_{n-p, r-1}
\\ = \frac{1}{r} \sum_{p=m}^k {n\choose p} a_{n-p, r-1}.$$
This yields
$$\bbox[5px,border:2px solid #00A000]{
a_{n,r} = \frac{1}{r}
\sum_{p=m}^{\min(n,k)} {n\choose p} a_{n-p, r-1}
\quad\text{where}\quad
a_{n,1} = [[m\le n\le k]].}$$
