Indistinguishable particles: obeing "Bose-Einstein Statistics" Let $r$ indistinguishable particles be placed at random into $n$ cells. If the particles obey "Bose-Einstein Statistics", prove that the probability that there are exactly $i$ particles in a given cell is 
$$\frac{{n+r-i-2 \choose r-i}}{{n+r-1 \choose r}}$$
I don't know Bose-Einstein Statistics principle. This troubles me to understand and solve the problem. I shall be obliged if some one be kind enough to solve me in full in details. I shall be highly benefited them.   
 A: The Bose-Einstein principle is that we consider the balls to be indistiguishable, and the boxes to be distinguishable, and we assume that the distinguishable arrangements are equally likely. 
The total number of possibilities (with these indistinguishable balls and distinguishable boxes) is $${ n+r-1 \choose r}.$$
This is easy to see. Picture, say, $r=5$ indistinguishable particles and $n=4$ cells. The $4$ cells can be represented by $3$ separating walls between the particles. (Think of $n-1$ separating walls in the general case.) If we represent both the walls and the balls by asterisks then we have the following picture $$********.$$ In order to get one arrangement of the $r=5$ particles in the $n=4$ cells we have to transform $r=5$ asterisks to balls, and the rest of the asterisks to walls:
$$\bullet\bullet|\bullet|\bullet\bullet|.$$
That is, the total number of the arrangements equals the number of possibilities to choose $r$ objects from a set of $n+r-1$ similar objects. One could reformulate this statement the following way: We have $n+r-1$ asterisks and we have to chose $n-1$ objets to play the role of the walls. As a result, we have $${ n+r-1 \choose r}={ n+r-1 \choose n-1}.$$
Furthermore, if we have an occupied cell with $i$ balls then the reamaining number of cells is $n-1$ and the number of the remaining particles is $r-i$. We can repeat the argumentation above and say that the number of arrangements with $i$ balls in a given cell is $${ n+r-i-2 \choose r-i}={ n+r-i-2 \choose n-2}.$$
The probability of the desirable event is then
$$\frac{{ n+r-i-2 \choose r-i}}{{ n+r-1 \choose r}}=\frac{{ n+r-i-2 \choose n-2}}{{ n+r-1 \choose n-1}}.$$
