Let \(A\) be an arrangement of hyperplanes in a real finite dimensional vector space \(V\). The components of the complement of the union of the hyperplanes are called the chambers of \(A\). The counting polynomial \(\sum_{i \geq 0} a_i t^i\) of \(A\) in a chamber \(C\) is defined by setting \(a_i\) equal to the number of chambers which are separated from \(C\) by exactly \(i\) hyperplanes. The set \(L(A)\) of all subspaces of \(V\) of the form \(H_1 \cap \dots \cap H_p\) with \(H_i \in A\), ordered by reverse inclusion, is a geometric lattice with least element \(V\) and greatest element \(T(A) = \bigcap_{H \in A} H\). An element \(Y \in L(A)\) is a complement of \(X \in L(A)\), if \(X \wedge Y = V\) and \(X \vee Y = T(A)\). Now \(A\) is called supersolvable, if there exists a maximal chain \(V = X_0 < X_1 < \dots < X_n\) in \(L(A)\), such that \(X_i\) is modular for \(i = 1,\dots,n\), which means that it does not admit two comparable complements. The purpose of the paper is to prove the following result: Let \(A\) be a supersolvable arrangement with a maximal chain as above. Let \(b_i\) be the number of hyperplanes of \(A\) which contain \(X_i\) and which do not contain \(X_{i-1}\) \((i = 1,\dots,n)\). Then there exists a chamber of \(A\) such that the counting polynomial of \(A\) in this chamber is equal to \(\prod^n_{i = 1} (1 + t + \dots + t^{b_i})\).