distributing z different objects among k people almost evenly We have z objects (all different), and we want to distribute them among k people ( k < = z ) so that the distribution is almost even.
i.e. the difference between the number of articles given to the person with maximum articles, and the one with minimum articles is at most 1.
We need to find the total number of ways in which this can be done
for example if there are 5 objects and 3 people the number of such ways should be 90.
I am not sure how we get this value.

(Edit: I just realize that this coincides with leonbloy's answer which has been questioned by the OP.)
Put $n:=\left\lfloor{z\over k}\right\rfloor$
and $r:=z-nk$. Then $r$ happy people receive $n+1$ objects and $k-r$ people $n$ of them. The $r$ happy people can be selected in $${k\choose r}$$ ways and will receive a total of $r(n+1)$ objects. These objects can be selected in $${z\choose  r(n+1)}$$ ways and can then be distributed among the $r$ happy people in $${\bigl(r(n+1)\bigr)!\over \bigl((n+1)!\bigr)^r}$$
ways. The remaining $(k-r)n$ objects can be distributed among the remaining $k-r$ people in
$${\bigl((k-r)n\bigr)!\over (n!)^{k-r}}$$
ways. The number $N$ you are looking for is given by the product of the displayed expressions:
$$N={k\choose r}{z\choose  r(n+1)}{\bigl(r(n+1)\bigr)!\over \bigl((n+1)!\bigr)^r}{\bigl((k-r)n\bigr)!\over (n!)^{k-r}}={k\choose r}{z!\over \bigl((n+1)!\bigr)^r (n!)^{k-r}} .$$
