“…. , 17,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16 But in this placement, there are two unused positions between 16 (the last appearance of an element of B 5 ) and 8 (the first appearance of an element of B 4 ), which can contain the elements 2 and 3 of B 4 . By placing these two elements immediately after 16, we obtain a solution that satisfies our assumptions, and is at least as good:…”