Integer partition of n into k parts recurrence

Let's consider an example: $n=8$ and $k=3$:

\begin{align*} \\ P(n,k)&=P(n-1,k-1)+P(n-k,k)&\\ \\ P(8,3)&=P(7,2)+P(5,3)&\\ \\ 5\quad&=\quad3\qquad+\quad2 \end{align*}

$$ $$

\begin{array}{rlrlrl} &P(8,3)\qquad=&& P(7,2)\qquad\qquad+&&P(5,3)\\ \hline\\ 8&=\color{red}{6+1}+1\qquad& 7&=6+1\\ &=\color{red}{5+2}+1\qquad&&=5+2\\ &=\color{red}{4+3}+1\qquad&&=4+3\\ &=\color{blue}{4+2+2}\qquad&&\qquad &\qquad5&=3+1+1\\ &=\color{blue}{3+3+2}\qquad&&\qquad&\qquad&=2+2+1 \end{array}

Note that the partitions of $P(8,3)$ that have smallest integer part equal to one correspond to the integer partitions of $P(7,2)$ whereas the partitions with smallest integer part $>1$ correspond to the partitions of $P(5,3)$: \begin{align*} 4+2+2=(3+1+1)+3\\ 3+3+2=(2+2+1)+3 \end{align*}

$p(n,k)$ is the number of ways to partition $n$ into $k$ parts. It is the same as the number of different ways of placing $n$ objects into $k$ pots. Firstly put $1$ object in each pot. Total $k$ objects have been put and now we have to put remaining $n-k$ objects into $k$ pots. Hence, $$ p(n,k)=p(n-k,1)+p(n-k,2)+\cdots+p(n-k,k-1)+p(n-k,k)$$ Also observe that, $$ p(n-1,k-1)=p(n-k,1)+p(n-k,2)+\cdots+p(n-k,k-1)$$ From the above two equations, we conclude: $$p(n,k)=p(n-1,k-1)+p(n-k,k)$$