Integer partition of n into k parts recurrence I was learning integer partition of a number n into k parts(with minimum 1 in each part) and came across this recurrence :
part(n,k) = part(n-1,k-1) + part(n-k,k) 
But, I cannot understand the logic behind this recurrence. Can someone please help me visualize this recurrence?
 A: I like to think of it as 'counting partitions with a certain property'...
We split the partitions $p(n,k)$ into $2$ disjoint classes that together comprise $p(n,k)$. So if we can count each class, then we've counted $p(n,k)$.
Let class $A$ contains all the partitions of $n$ into $k$ parts where at least $1$ of the partition elements is a $1$. So these all look like $1 +$ stuff.
The rest, class $B$, don't have any $1$ elements. These partitions could look like almost anything, but you won't find a $1$ in there anywhere. There are still $k$ parts that add to $n$.
It should be clear that these two classes are disjoint and taken together comprise all the partitions of $n$ into $k$ parts.
So if we count class $A$ and class $B$ we've counted all the partitions and we'll have our answer $p(n,k)$, so let's do that ...
Look at class $A$. We already noticed every partition in class $A$ looks like $1 +$ stuff. Well, the sum of stuff is $n-1$. How many pieces are in stuff? They all have $k-1$ pieces. There are no other restrictions on stuff, but then stuff is just all the partitions of $n-1$ into $k-1$ parts. So the count of class $A$ is $p(n-1,k-1)$.
Now look at class $B$. The sum of the parts is still $n$. There are $k$ parts, but each part is $\geq 2$. That means we can subtract $1$ from each part, but then the sum is $n-k$. Each part is now $\geq 1$, but that's just the definition of the partition of $n-k$ into $k$ parts. So the count of class $B$ is $p(n-k,k)$.
That's it.
A: $p(n,k)$ 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:
$$p(n,k)=p(n-1,k-1)+p(n-k,k)$$
A: 
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*}
