Question

A box is divided into $m$ equal compartments into which $n$ balls are thrown at random; find the probability that there will be $p$ compartments each containing $a$ balls, $q$ compartments each containing $b$ balls, $r$ compartments each containing $c$ balls, and so on, where $pa+qb+rc+.....=n$.

Answer 1

Since each of the $n$ balls can fall into any one of the $m$ compartments the total number of cases which can occur is $m^n$, and these are all equally likely.

To determine the number of favourable cases we must find the number of ways in which the $n$ balls can be divided into $p,q,r,....$ parcels containing $a,b,c,...$ balls respectively.
First let's choose any $s$ of the compartments, where $s$ stands for $p+q+r+....;$
The number of ways in which this can be done is $\frac{\lfloor m}{\lfloor s\lfloor m-s}$.............$\left(1\right)$.
Next subdivided the $s$ compartments into groups containing $p,q,r,....$ severally, the number of ways in which this can be done is
$\frac{\lfloor s}{\lfloor p\lfloor q\lfloor r}$..............$\left(2\right)$.
Lastly, distribute the $n$ balls into the compartments, putting $a$ into each of the group of $p,$ then $b$ into each of the group of $q$ and $c$ into each of the group of $r,$ and so on. The number of ways in which this can be done is $\frac{\lfloor n}{\left(\lfloor a{)}^p\right(\lfloor b{)}^q(\lfloor c{)}^r..........}$..................$\left(3\right)$.
Hence, the number of ways in which the balls can be arranged to satisfy the required conditions is given by the product of the expressions $\left(1\right),\left(2\right),\left(3\right)$. Therefore the required probability is
$\frac{\lfloor m\lfloor n}{m^n\left(\lfloor a{)}^p\right(\lfloor b{)}^q(\lfloor c{)}^r...........\lfloor p\lfloor q\lfloor r............\lfloor m-p-q-r-...}$.