DanA> As I see it, there's only one natural way to extend the binomial symbol n_C_k to almost all reals (and also complexes), and that is via the gamma function 𐅃(z), since for positive integers z we have from 𐅃(z) = (z-1)! that (*) n_C_k = 𐅃(n+1) / (𐅃(k+1) 𐅃(n-k+1)) This definition would imply that 0_C_0 is equal to 1, and also that for integer n and k then for either a) n >= 0 AND (k <= -1 OR k >= n+1) we have n_C_k = 0; OR b) n <= -1 AND (k <= -1 AND k >= n+1), --or more simply-- n+1 <= k <= -1 we have n_C_k = 0. (Since, the two simple poles in the denominator force the quotient to equal 0, despite the simple pole in the numerator.) Since the gamma function never takes the value 0 and has poles only at the nonpositive integers, a) and b) should cover all cases where the definition (*) of n_C_k = 0. --Dan Why settle for almost all? Using z! := Gamma[z+1], for the 3rd time, _nC_m := (n choose m) := Limit[Limit[x!/(y!*(x-y)!),y->m],x->n] covers *all* the cases. Picture: Overlay on the symmetrical surface x!/(y!*(x-y)!) a symmetrical grid of arrows pointing (unsymmetrically) leftward at each gridpoint, indicating that if we always approach from the *right*, we always get the *right* answer. JPropp> I can't remember the full story, but here's a summary of part of it (which I learned about from Ira Gessel): Assume that we can define C(m,n) for all integers m and n so that C(m,n)=(-1)^n C(-m+n-1,n) and C(m,n)=C(m,m-n) for all m and n. Then applying these two transformations alternately three times we get C(m,n) = -C(m,n): C(m,n) = (-1)^n C(-m+n-1,n) = (-1)^n C(-m+n-1,-m-1) = (-1)^(m+n+1) C(-n-1,-m-1)= (-1)^(m+n+1) C(-n-1,-n+m) = -C(m,-n+m) = -C(m,n). Jim Propp OK, that just says we can't define C(m,n) for all integers m and n so that C(m,n)=C(m,m-n). Good! (I'm surprised Ira, a CAS guy, would believe C(m,n)=C(m,m-n) for all m and n. Knuth Vol 1 must clarify this.) What does Maple say? Fred> I understand now what Bill is driving at. The conclusion seems to be that the Gamma (or Beta) function definition is essentially useless for defining n_C_m for integers m, n < 0 . rwg> Huh? That limit I keep repeating seems to do it all. Fred> It looks like fans of the radiation symbol option need to come up with some convincing alternative justification for taking (*) as an axiom, to extend to the region m < 0, n < 0 . At any rate when (n,m) = (-1,-1) , there are indeed situations where the value 1 gives a consistent combinatorial result; I'm not sure though whether there may be others where 0 is preferable. Fred Lunnon I'd be interested to see a situation where (-1 Choose -1) := 1 avoids absurdity, barring reckless use of C(m,n)=C(m,m-n)). The radiation symbol is more toxic than radiation. This is (n choose m): 1 1 -5 1 -4 1 -3 6 1 -2 3 1 -1 1 -1 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 This biwedge extension is typical of many number triangles, including Stirling and Eulerian. --rwg