So, you are interested in f(n,k)=sumki=0(−1)ibinomkibinom2n−kn−i.
Simple manipulations show f(n,k)=frack!(2n−k)!(n!)2left[sumni=0(−1)ibinomnibinomnk−iright]
Now the second factor counts the coefficient of xk in (1−x2)n and therefore if k is odd f=0 otherwise f=(−1)frack2frack!(2n−k)!n!(k/2)!(n−k/2)! which is far from zero...
EDIT: On a different note I see the result is a signed generalized Catalan number of degree 2 (I was not aware they satisfied such simple identities). Since usually providing combinatorial interpretations for generalized Catalan numbers is not easy, may I ask in what combinatorial context did you face the above calculation?
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