We suppose $n$ and $p$ are two positive integers.
A) In how many ways can you divide $p$ identical envelopes in $n$ mailboxes? (Each mailbox can hold several envelopes at the same time)
B) In how many ways can you divide $p$ distinct envelopes in $n$ mailboxes?
C) In how many ways can you divide $p$ identical envelopes in $n$ mailboxes, such that no mailbox remains empty? (it is assumed that $p$ is greater than or equal to $n$)
Any thoughts?
A and C) If you have $n$ different mailboxes, let $x_i$ be the number of envelopes in the box $i$. The number of ways you can put $m$ identical envelopes in $n$ boxes is the number of solutions to following equation: $$ x_1+x_2+...+x_n=m $$ If $x_i>0$, the number of ways is $\binom{m-1}{n-1}$, otherwise if $x_i\geq 0$, the number of ways is $\binom{n+m-1}{n-1}$.
Proof: When $x_i>0$, Each case of putting envelopes in mailboxes can be represented by a sequence of $m$ number of $A$ and $n-1$ number of $B$ such that there is no two neighbor $B$s. For instance if we have $m=5$ envelopes and $n=3$ mailboxes, and if there are respectively $1,1,3$ in each box we can represent in by: $$ A\,\,\,\,\,B\,\,\,\,\,A\,\,\,\,\,B\,\,\,\,\, AAA $$ In this representation, the problem is equivalent to putting each $B$ only in one of the $m-1$ places between two consecutive $A$s (No $B$ in the beginning and the end of the sequence) and no two $B$s in the same place, therefore we have to choose $n-1$ places from $m-1$. When $x_i\geq 0$, it is enough to replace $x_i=y_i-1$ and the problem turns to the previous one.
B) If you have $m$ distinct envelope and $n$ different boxes, then you have multiply the previous numbers by $m!$.
Proof: The proof essentially follows the same idea with the difference that $A$s are distinct and there are $m!$ different ways to put them as a sequence.