A:=(n,k)->(n!/k!)*binomial(k,n-k)*2^(2*k-n);
convert(A(n,k), factorial);
B:=(n,k)->n!*2^(2*k-n)/((n-k)!*(2*k-n)!);
A(1,0); # gives 0
B(1,0); # Error, div by zero
Peter
To work around that weakness here one can use the limit to get the result:
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