Gibbs free energy and entropy in enzyme catalysis
As @canadianer hinted, the answer is enthalpy.

ΔG = ΔH - TΔS

For binding to be a spontaneous process ΔG must be negative. If the binding of a ligand to a protein decreases system entropy then ΔS is negative and the entropy term -TΔS is positive. However ligand binding involves the formation of various protein-ligand interactions (hydrogen bonds, ionic interactions, van der Waals interactions) all of which make a negative contribution to the change in enthalpy, so ΔH is negative. If |ΔH| > |TΔS| then ΔG will be negative. 

This simplified account ignores the parallel changes taking place due to the ligand 'dissociating' from the solvent.