# Why does Multinomial[-1/2, -1/2, 1] give Indeterminate rather than 1/Pi?

GROUPS:
 Is this a bug? Or there might be something I don't understand. In[1]:= Multinomial[-1/2, -1/2, 1] "During evaluation of In[1]:= Infinity::indet: Indeterminate expression 0 ComplexInfinity encountered." Out[1]= Indeterminate None of the arguments are located at poles of the factorial function (negative integers), so I expected this answer: In[2]:= (-1/2 - 1/2 + 1)! / ((-1/2)! (-1/2)! (1)!) Out[2]= 1/Pi Checking with Simplify using Assumptions: In[3]:= Assuming[a == -1/2 && b == -1/2 && c == 1, Simplify[ Multinomial[a, b, c] == 1/Pi ] ] Out[3]= True Thanks! (I'm running 11.2) In[4]:= \$Version Out[4]= "11.2.0 for Mac OS X x86 (64-bit) (September 11, 2017)" 
 Michael Rogers 1 Vote It seems it might be using the identity Multinomial[a, b, c] == Binomial[a + b, b] Binomial[a + b + c, c] The identity is only generically true, but Indeterminate for a = b = -1/2 and c = 1.A possible workaround: Multinomial[a, -1/2, 1] /. a -> -1/2 
 Mariusz Iwaniuk 1 Vote A possible workaround: Binomial[a + b, b]*Binomial[a + b + c, c] // FunctionExpand // FullSimplify (* Gamma[1 + a + b + c]/(Gamma[1 + a] Gamma[1 + b] Gamma[1 + c])*) as a function: MultinomialUSER[a_, b_, c_] := Gamma[1 + a + b + c]/(Gamma[1 + a] Gamma[1 + b] Gamma[1 + c]) MultinomialUSER[-1/2, -1/2, 1] (* 1/\[Pi] *)