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.
 A: Let $k_1$ be the number of people that receive $n_1 = \lfloor \frac{z}{k}\rfloor $ objects, $k_2$ receive $n_2= n_1 +1$ objects. Clearly, $k_2 = z \bmod k$, $k_1+k_2=k$, $k_1 n_1 + k_2 
n_2 = z$.
Assuming the objects are distinguishable, the total number of ways is
$$ {k \choose k_1} \frac{z!}{(n_1!)^{k_1} (n_2!)^{k_2}}  $$
A: Each person will get either $\lfloor \frac{z}{k}\rfloor$ or $\lceil \frac{z}{k}\rceil$ objects.  These are the floor and ceiling functions.
A: As k is always less than z; 
First choose k elements from z; distribute it one for each person.
Then the remaining objects for your condition to be true;
Lets call a(1) = z-k;
Again repeat the process [n is the number of cycles]
till a(n) = z-n*k < k
Then distribute the remaining; Randomly. one person will only get one
This way the difference between the number of articles given to the person with maximum articles, and the one with minimum articles is at most 1.
A: (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:
$$N={k\choose r}{z\choose  r(n+1)}{\bigl(r(n+1)\bigr)!\over \bigl((n+1)!\bigr)^r}{\bigl((k-r)n\bigr)!\over (n!)^{k-r}}={k\choose r}{z!\over \bigl((n+1)!\bigr)^r (n!)^{k-r}} .$$
