what happens when in a combination, $n \choose k$, $n < k$? Is this 0? What happens when in a combination, $n \choose k$, $n < k$? Does this result in a zero, or is this undefined?
 A: Not necessarily $0$. We have definition
$$\binom{n}{k} = \frac{n(n-1)...(n-k+1)}{k!}$$
So for example if we take $n = -3$ and $k = 2$, we have
$$\binom{-3}{2} = \frac{(-3)\cdot(-4)}{2!} = 6$$
If you don't believe, see https://www.wolframalpha.com/input/?i=C(-3,2). Selection of objects is just an analogy for natural number values of $n$ and $k$, I suggest you to use the definition instead, for a more general result.
A: Hint:

Since $\binom{n}{k}$ is the the number of ways you can choose a subset of $k$ numbers out of $n$, in how many ways can you do such a choice when $n<k$?

P.S. Ask for further explanation, if needed.
A: Imagine this intuitively. The value: $$\binom{n} {k} $$ deals with the number of ways of choosing $k$ objects from $n$ objects. 
Now, imagine $n<k$. How many ways of selecting $k$ objects are possible? 

 Yes, the answer is zero. 

A: For natural $k,$ the $k^{\text{th}}$ degree polynomial $\binom xk=\frac{x(x-1)\cdots(x-k+1)}{k!}$ is $0$ only for $x=0,1,\dots,k-1.$
You can easily verify the identity
$$\binom{-x}k=(-1)^k\binom{x+k-1}k.$$
For example, $\binom{-2}n=(-1)^n\binom{n+1}n=(-1)^n(n+1).$
Thus the case $m=-2$ of the binomial theorem
$$(1+x)^m=\sum_{n=0}^\infty\binom mnx^n$$
simplifies to
$$(1+x)^{-2}=\sum_{n=0}^\infty\binom{-2}nx^n=\sum_{n=0}^\infty(-1)^n(n+1)x^n=1-2x+3x^2-4x^3+\cdots$$
When $x$ is a natural number, the polynomial $\binom xk$ is equal to the number of $k$-element subsets of an $x$-element set. This can be used to give bijective proofs (proofs by counting) of many combinatorial identities, such as Pascal's identity
$$\binom xk=\binom{x-1}k+\binom{x-1}{k-1}\ \ (k=1,2,3,\dots).$$
Namely, you can prove it for natural values of $x$ by a counting argument, and then since both sides are polynomials you can conclude that the identity is valid for all values of $x.$
A: We have $n \choose r$. (The common notation is $n$ and $r$)
\begin{align}
n \choose r & = \frac{n!}{(n-r)!r!}\\
\end{align}
$n \lt r \implies n-r < 0$. The factorial is not defined for negative numbers, so $n \choose r$ does not have an answer if $n < r$.
