Cell and ball probability problem n distribution of balls are placed at random in n cells. Find the probability that.
a) First cell will contain "h" specific balls.
b) The occupancy number of the first two cells are "h1" and "h2" specific balls.
there 1 and 2 are suffix of h.
for a tried like this. first I chose h balls from the balls and put it in the first cell and the n-h balls have to be put in n-1 cells I thought the method of positive integral solution method but am not sure about it. and for the second one I am completely clueless how to start.
 A: The probability that $h_1$ specified balls land in Cell $1$, along with possibly others,  is $\left(\frac{1}{n}\right)^h$. For each of the specified balls has probability $\frac{1}{n}$ of landing in Cell $1$, and what happens to Ball $i$ is independent of what happens to Ball $j$ if $i\ne j$. 
If we want only these specified balls to fall in Cell $1$, we  have to multiply $\left(\frac{1}{n}\right)^h$ by $\left(\frac{n-1}{n}\right)^{n-h}$.
Given that exactly $h_1$ specified balls land in Cell $1$, the probability $h_2$ specified balls land in Cell $2$, and the rest elsewhere, is
$$\left(\frac{1}{n-1}\right)^{h_2} \left(\frac{n-2}{n-1}\right)^{n-h_1-h_2}.$$
So the probability that a specified $h_1$ land in Cell $1$, and a specified $h_2$ land in Cell $2$ is
$$ \left(\frac{1}{n}\right)^{h_1} \left(\frac{n-1}{n}\right)^{n-h_1}\left(\frac{1}{n-1}\right)^{h_2} \left(\frac{n-2}{n-1}\right)^{n-h_1-h_2}.$$

More useful in applications is when the balls are not specified. For the probability, one multiplies the probability above by $\binom{n}{h_1}\binom{n-h_1}{h_2}$. Some simplification can be made.
