Counting Graphs This problem actually relates to multi processing computer architecture, but boils down to a mathematical expression quite difficult to understand.
The image below explains the problem and the last expression (N-1)C(r-1) is some thing I am NOT able to understand.

I took an example of 9 tasks and 3 levels so that N=9 and r=3.
Going by the description at the bottom, each level at least should have one task and hence division of rest of 6 tasks in three levels counts the number of process graphs. I did try as below : 
Enumerate in how many ways number 6 be partitioned in three places : 
2,2,2  <-- Only 1 way to express so that it totals to 6
1,1,4  <-- Only 3 ways to express so that it totals to 6
3,2,1  <-- 6 ways to express so that it totals to 6
4,2,0  <-- 6 ways to express so that it totals to 6
5,1,0  <-- 6 ways to express so that it totals to 6
6,0,0  <-- Only 3 ways to express so that it totals to 6
3,3,0  <-- Only 3 ways to express so that it totals to 6
So in all this totals to 1+3+6+6+6+3+3 = 28 ways
Going by the expression the value should be 8C2 which evaluates to 28.
However, what I lack is to understand, how we have arrived at the result (N-1)C(r-1).
Thanks for reading. 
 A: A standard way of counting arrangements like these is the so-called "stars and bars" argument.  If you have a way of splitting a total of $m$ into a sum of $r$ parts, then that's the same as a way of listing $m$ stars ($\star$) separated by $r-1$ bars ($|$).  For example:
$$\begin{align*}
6 = 2+2+2 &\implies \star\star|\star\star\ |\star\star\\
6 = 5+0+1 &\implies \star\star\star\star\star\ |\ |\star
\end{align*}$$
How many ways are there to separate $m$ stars by $r-1$ bars?  Well, that's a total of $m + (r-1)$ symbols, and you're counting the ways for $(r-1)$ of them to be bars, so there are ${m + (r-1) \choose r-1}$ ways total.
In your case, $m=N-r$, so it simplifies to $${(N-r) + (r-1)\choose r-1} = {N-1\choose r-1}\text{ ways.}$$
On the other hand, you can count directly by imagining that instead of trying to fit $N$ tasks into $r$ preexisting levels, the tasks come with an ordering and you are deciding which of the tasks are the last tasks in each level.  Since there are $r$ levels, you have to choose $r$ of the $N$ tasks to be level-finishing tasks.  But the final task has to be the $r$th level-finisher, so you are really only choosing which of the remaining $N-1$ tasks finish the first $r-1$ levels.  There are ${N-1\choose r-1}$ ways to choose.
