Is $^nC_r$ defined when $r>n$? I want to know if $^nC_r$ is not defined when $r>n$ or is it just equal to 0?
By its definition, we know that, $^nC_r$ is the number of ways of selecting $r$ things out of $n$ distinct things, in this manner, $^nC_r$ should be equal to 0.
But, by looking on its mathematical statement, i.e., $^nC_r$ = $\frac{n!}{r!(n-r)!}$, so if $r>n$, then $(n-r)!$ will not be defined resulting in $^nC_r$ being undefined.
So, What is the correct way to look at this?
 A: By convention, $_{n}C_{r}$ is defined to be zero if $r>n$ or if $r$ is negative. This is so that the rule for generating Pascal's triangle would still hold with this extended definition.
A: I can't think of anything concrete at the moment, but yes, frequently the idea of $^nC_r$ is extended to $r > n$ by setting it to be $0$. You're right that $^nC_r$ is, typcially, undefined when $r > n$, but the same could once have been said about $\sin(x)$ where $x > \pi/2$ or $x < 0$.
While it doesn't strictly make sense with the factorial formula, as you noted, it does make sense with the combinatorial definition of $^nC_r$. However, if you'll tolerate some pseudo-logic here, recall that, for $n \ge 1$,
$$\frac{1}{(n-1)!} = \frac{n}{n!}.$$
If we try to force this to make sense when $n = 0$, we get
$$\frac{1}{(-1)!} = \frac{0}{0!} = \frac{0}{1} = 0.$$
Of course, no real number satisfies this! You could actually make this more rigorous by appealing to the holomorphic function $\frac{1}{{\Gamma}(z)}$, and removing its discontinuities.
A: There is of course no conventional  combinatorial interpretation of something like $^3C_5$ (as far as I know). However, there is an algebraic interpretation which makes the convention $^nC_r=0$ for $r>n$ natural: in particular, switching notation from $^nC_r$ to $\binom{n}{r}$, we have 
$$(x+y)^n=\sum_{r=0}^n\binom{n}{r}x^ry^{n-r}.$$ This is, of course, the binomial theorem. In particular, we have that $$(x+1)^n=\sum_{r=0}^n\binom{n}{r}x^r.$$ In many situations, particularly when dealing with generating functions, it makes sense to want to think of a polynomial as the series $\sum_{r\geq0}a_rx^r$ (this is the general form of an ordinary generating function), where all but finitely many of the $a_r$ are zero. We want to allow ourselves to write $$(x+1)^n=\sum_{r\geq0}\binom{n}{r}x^r,$$ which is very notationally convenient, but we can only do this by accepting that $\binom{n}{r}$ is $0$ for $r>n$. Indeed, in algebraic contexts at least, this is the usual convention. 
A: The binomial coefficient $^n C_r$ is defined as the number of ways in which one can choose $r$ unordered possibilities from $n$ options (that is the definition for integer $r$ and $n$). There are zero ways to pick $r$ things from $n$ things if $r > n$. For this reason, we put $^n C_r = 0$ for $r > n$.
A: You can use the alternate formula
$$^n C_r=\frac{n(n-1)\ldots (n-r+1)}{r!}$$
for $r\ge 0$ integer.
