6
$\begingroup$

I am working on parameter estimation and one of the estimators involves a summation of $_nC_k$ ($n$ choose $k$) expressions. For some iterations, I need to compute expressions like $_0C_1$, $_0C_2$, etc.

In general how do we compute $_nC_k$ when $n$ is less than $k$? Do we still use the formula $\frac{n!}{(n-k)!k!}$ and use the gamma function to compute the negative factorial?

Thanks!

$\endgroup$
  • 13
    $\begingroup$ By convention $\binom{n}k=0$ when $n<k$. $\endgroup$ – Brian M. Scott Dec 30 '12 at 14:16
  • $\begingroup$ I take it $n$ is still a non-negative integer? $\endgroup$ – Gerry Myerson Dec 30 '12 at 14:20
  • $\begingroup$ Note that you also can't "use the gamma function to compute negative factorial[s]" to get an answer, since the function is not defined on the non-positive integers. $\endgroup$ – TMM Dec 30 '12 at 14:26
  • $\begingroup$ @TMM: Use the (entire) reciprocal gamma function instead, then: $$\binom nk = \frac{\frac{1}{\Gamma(n-k+1)} \frac{1}{\Gamma(k+1)}}{\frac{1}{\Gamma(n+1)}}$$ $\endgroup$ – hmakholm left over Monica Dec 30 '12 at 15:02
6
$\begingroup$

You could define $$_nC_k \text{ or } {n \choose k} = \frac{n(n-1)\cdots(n-k+2)(n-k+1)}{k(k-1)\cdots 2\cdot 1}$$ for positive integer $k$, and ${n \choose 0} = 1$ as the quotient of empty products.

This would give the usual values for non-negative integer $n \ge k$.

It would also give values for other real $n$. For non-negative integer $n \lt k$ including ${0 \choose k}$ it would give $0$ as the numerator involves multiplication by $0$ while the denominator does not.

$\endgroup$
  • $\begingroup$ I see. Thanks @Henry and Brian M. Scott and to the others who responded!!! $\endgroup$ – Tomas Jorovic Dec 30 '12 at 23:06

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.