The reference mentions that this identity is from combinatorics: that is, it counts things.  

What does it count?  Consider $N$ objects.  Once and for all, divide those $n$ things into a group of $K$ of them, which I will call "red," and the remainder, which I will call "blue."  Each subset of $n$ such objects determines, and is determined by, its red objects and its blue objects.  The number of such sets with $k$ red objects (and therefore $n-k$ blue objects) equals the number of ways to choose $k$ red objects from all $K$ ones (written $\color{red}{\binom{K}{k}}$) times the number of ways to choose the remaining $n-k$ blue objects from all the $N-K$ ones (written $\color{blue}{\binom{N-K}{n-k}}$).

Now if $k$ is not between $0$ and $K$, then there is no $k$-element subset of $K$ things, so $\binom{K}{k}=0$ in such cases.  Similarly, $\binom{N-K}{n-k}=0$ if $n-k$ is not between $0$ and $N-K$.  (This not only makes sense, it is actually how good software will evaluate these quantities.  Ask R, for instance, to compute choose(5,6) or choose(5,-1): it will return the correct value of $0$ in both cases.)

Summing over all possible numbers $k$ shows that 

$$\binom{N}{n} = \sum_k \color{red}{\binom{K}{k}}\color{blue}{\binom{N-K}{n-k}}$$

and as you read this you should say to yourself "any $n$ objects are comprised of some number $k$ of red objects and the remaining $n-k$ blue objects."

The sum needs to include all $k$ for which both the terms $\color{red}{\binom{K}{k}}$ and $\color{blue}{\binom{N-K}{n-k}}$ are nonzero, but it's fine to include any other values of $k$ because they will just introduce some extra zeros into the sum, which does not change it.  We just need to make sure all relevant $k$ are included. It suffices to find an obvious lower bound for it ($0$ will do nicely and is more practicable than $-\infty$!) and an obvious upper bound ($N$ works because we cannot find more than $N$ objects altogether).  A slightly better upper bound is $n$ (because $k$ counts the red objects in a set of $n$ things).  Thus, writing these bounds explicitly and dividing both sides by $\binom{N}{n}$, we obtain

$$1 = \sum_{0\le k\le n}\frac{\color{red}{\binom{K}{k}}\color{blue}{\binom{N-K}{n-k}}}{\binom{N}{n}} .$$

Despite the notation, this formula does not implicitly assert that all values of $k$ in the range from $0$ to $n$ can occur in this distribution.  About the only reason to fiddle with the inequalities and figure out what the smallest possible range of $k$ can be would be for writing programs that loop over these values: that might save a little time adding up some zeros.

How the hypergeometric distribution sums to 1?
In my ongoing quest to interface with some legacy equipment, I've discovered that the vendor supplied software sets the special characters to:

00 00 00 00 11 13 


But the SerialPort class of .NET set them to:

1A 00 00 1A 11 13 


I suppose that I have two questions:


What do these bytes mean?
How can I tell SerialPort to use a specific set of special characters?


The latter is all I really care about, but I suspect the former is going to be useful to know.



Update: The following doesn't work:

byte[] specialchars = {
    0x00,
    0x00,
    0x00,
    0x00,
    0x11,
    0x13
};

this.port.NewLine = System.Text.Encoding.ASCII.GetString(specialchars);




Update 2: As requested, here are Portmon logs for the vendor supplied app (filtered to remove the many thousands of IOCTL_SERIAL_GET_COMMSTATUS entries) and for my attempt to match even the first exchange.
