4
$\begingroup$

Would it be reasonable to define $\binom{n}{n+1} = 0$?

My thinking is that it should be possible, since there are no ways to select $n+1$ items from a group of $n$ objects.

$\endgroup$
2
  • 2
    $\begingroup$ There are some schools of thought that do define such a combination to equal 0. $\endgroup$ Nov 9 '18 at 16:59
  • 4
    $\begingroup$ Yes, Pascal's triangle is "by default" zero outside the range $[0,n]$ In row $n$. Likewise, $\binom n{-1}=0$. $\endgroup$
    – user65203
    Nov 9 '18 at 17:00
5
$\begingroup$

One of many equivalent definitions of Newton symbol is that $${n \choose k} = \frac{n\cdot(n-1)\cdot(n-2)\cdot\ldots\cdot(n-k+1)}{k!}.$$ Therefore $${n \choose n+1} = \frac{n\cdot(n-1)\cdot(n-2)\cdot\ldots\cdot1\cdot 0 }{(n+1)!} = 0.$$

$\endgroup$
0
0
$\begingroup$

The only sensible definition is zero. This can be formalised using the Gamma function; note $1/\Gamma(z)=0$ for $n=0,-1,-2,\ldots$.

$\endgroup$
0
$\begingroup$

Think this as number of ways to select $n+1$ balls form $n$ balls = $\boxed{0}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.