The different ways of stabbing disjoint convex sets (Q1184163)

Let \(A\) be a family of sets in \(\mathbb{R}^ d\). A straight line that intersects every member of \(A\) is called the common transversal of \(A\). If sets of \(A\) are convex and pairwise disjoint then a common transversal meets the members of \(A\) in a definite order. The transversal thus determines two permutations of \(A\), one being the reverse to the other. This pair of permutations is called a geometric permutation of \(A\). The authors continue the study of the following problem: How many geometrical permutations may have a family of \(n\) disjoint convex sets in \(\mathbb{R}^ d\)? Main results: 1. A family of \(n\) pairwise disjoint convex sets in \(\mathbb{R}^ d\) having \((n/(d-1))^{d-1}\) geometric permutations is constructed. 2. A family of \((d+1)\) translates in \(\mathbb{R}^ d\) having \((d+1)!/2\) geometric permutations is constructed.