I tell the beginners that $(-n)!= \pm \infty$, if $n \in N$. Next, I tell them that the definition of ${\nu \choose k}$ as $$ \frac{\nu (\nu-1) (\nu-2) (\nu -3) ...(\nu-k+1)}{k!}~~~~(1)$$ is the most practical one where $\nu$ need not be necessarily a positive integer. More generally it could be a negative integer, fraction or even a complex number. This helps them in understanding that why ${n \choose -1}=0$. Books keep such coefficients under the rug. Eq. (1) helps them to calculate ${-3 \choose 2}=\frac{-3.(-3-1)}{2}=6,{-5 \choose 3}= \frac{-5. (-5-1).(-5-2)}{3!}=-35$, ${1/2 \choose 3} =\frac{(1/2)(1/2-1)(1/2-2)}{3!}=\frac{1}{16}$ and ${-1/2 \choose 4}=\frac{(-1/2)(-1/2-1)(-1/2-2)(-1/2-3)}{4!}=\frac{35}{128}$ with ease and confidence. If they insist on using $${n \choose r}=\frac{n!}{(n-r)! ~r!},\quad n,r \in I~~~~(2)$$ for calculating ${-4 \choose 3}$. I further tell them that $(-4)!=-\infty....-10.-9.-8.-7.-6.--6.-5.-4$ and $(-7)!=-\infty....-10.-9.-8.-7$, so that $${-4 \choose 3} =\frac {(-4)!}{(-7)!~ 4!}=\frac{(-\infty....-10.-9.-8.-7).-6.-5.-4}{(-\infty....-10.-9.-8.-7)~ 1.2.3}=-20.$$ They also understand from (2) as to why ${4 \choose 7}$ is zero, also because ${4 \choose 7}= \frac{4!}{7! (-3)!}=\frac{1}{\infty}=0$. I also tell them the usual identity that $${-r \choose k}=(-1)^k {r+k-1 \choose k} ~~~~(3)$$ They find this one suspiciously interesting. Nonetheless, they use (1,2) to check that ${-4 \choose 3}=-{6 \choose 3} =-20$ to have faith in (3).They also find it delightful to check that $(1-x)^{-1}$ is both an infinite geometric progression and a binomial series as well by realizing that all the binomial coefficient are of the type ${-1 \choose k}$ and these are nothing but $(-1)^k$.

A few days back I told some of these things to one starter on the Mathstackexchange in reply to his question. I am afraid! he didn't get back to me and also he seems to be uninterested in his question itself. So I decided to tell these things loudly for your nod. I believe the good ones explore such things themselves, but others need to be told. Remember, in the beginning we are told that anything multiplied by zero is zero. But later we learn that anything should not be $\infty$ and we learn the indeterminate form like $ 0 \times \infty$.

I welcome your comments and suggestions in this KinderGarten of binomial coefficients..


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  • $\begingroup$ I think just calculating ${-r\choose k}$ doesn't make sense. It should be included that they form probability distribution of certain types. $\endgroup$ – Archis Welankar Jun 14 at 12:30
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    $\begingroup$ Maybe matheducators.stackexchange.com could be more appropriate for your question? $\endgroup$ – trancelocation Jun 14 at 12:32
  • $\begingroup$ @Archis Welankar These coefficient are required in the expansions such as $(1-x)^{-4}$, in this one would require ${-4 \choose k}$. $\endgroup$ – Dr Zafar Ahmed DSc Jun 14 at 12:35
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    $\begingroup$ I wouldn't use infinite products like that in a definition. There is an equally good argument in favor of $1\cdot 2\cdot 3\cdot\cdots\cdot $ being $0$ (because what other integer is divisible by every positive integer?). But it's a nice mnemonic if nothing else. $\endgroup$ – darij grinberg Jun 14 at 13:48
  • $\begingroup$ @darij grinberg Thanks for your comment. $\endgroup$ – Dr Zafar Ahmed DSc Jun 14 at 14:02