# value if $\binom{n}{c}$ when $n$ is a positive integer and $r$ is a negative integer

The formula we use for ncr is $\frac{n!}{(n-r)!r!}$

for example, $\binom{5}{2}= \frac{5!}{2!3!} = 10$

Special case is also there $\binom{n}{0}=1$. As per the formula, $\binom{n}{0} = \frac{n!}{0!n!}$.

I know cases like $\binom{n}{-1}$ (i.e., when $r$ is negative) does not make sense as selection of negative quantity from a finite quantity. But, when trying for a proof in my physics exercises, I got intermediate results like $\binom{\,2}{-1}$.

when $n$ is a positive integer and $r$ is a negative integer, can we take $\binom{n}{c}$ as zero?

• Yes: $\binom{n}k$ is defined to be $0$ when $k>n$ or $k<0$. – Brian M. Scott Dec 22 '15 at 20:58
• @BrianM.Scott, thanks. But when I use the formula, I get factorial of a negative number, for example, consider 5C(-1). as per the formula, it is 5!/(-1!* 6!). So the formula is not applicable in such cases and by definition it is zero, is it? – Kiran Dec 22 '15 at 21:00
• @Kiran See mathworld.wolfram.com/BinomialCoefficient.html -- in particular, (5) – Marcus Andrews Dec 22 '15 at 21:03
• thanks for the info, I was aware of only the basic formula. Thanks again, – Kiran Dec 22 '15 at 21:04
• That being said, if you're not familiar with this, theres a decent chance you're doing something wrong in your exercise. – Batman Dec 22 '15 at 21:16