Let $p^m(n, k)$ denote the number of partitions of $n$ having exactly $k$ parts with each part greater than or equal to $m$. Show that $p(n-k, k)=p^2(n, k)$, with the convention that $p(n, k)=0$ if $n<k$.
$p(n-k,k)$ denote the number of partitions of $n-k$ into exactly $k$ parts. I am struggling with this exercise. Any hint?
Imagine you have $n$ items and you want to divide them into $k$ parts, and each part must have more than or equal to two items.
First, you take $k$ items out and place them aside, and then partition the rest $n-k$ items into $k$ parts and make each part have more than or equal to one item (notice that this process is exactly $p(n-k, k)$). Then remember the $k$ items you put aside before, and put each one of them into each of the group. Then each one of your group will have more than or equal to $2$ items. Notice that this is exactly $p^2(n, k)$. So $p(n-k, k) = p^2(n,k)$.