1

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?

5 Answers5

1

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.

1

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.

user744868
  • 1,533
1

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.

YiFan Tey
  • 17,431
  • 4
  • 28
  • 66
1

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 $r > n$.

0

You can use the alternate formula $$^n C_r=\frac{n(n-1)\ldots (n-r+1)}{r!}$$ for $r\ge 0$ integer.

orangeskid
  • 53,909