Assume that L(A) is non-empty. If A accepts a word of length >= n, there is an accepting run of A that has at least one loop. If we cut out all the loops in this run, we get an accepting run (on some other word) that visits each of the n states at most once, and hence there are at most n-1 transitions, meaning there is a word in L(A) of length at most n-1. Thus the shortest word in L(A) has length at most n-1.

The predecessor of A is the rightmost node in the left subtree of A.

Let us say a person's profile is a function from the set of all arrangements to {0,1} (indicating whether a particular arrangement is liked by that person or not). There are 2⁵ possible arrangements. Thus there are 2²⁵ possible profiles. Two people with the same profile have the same likes and dislikes, and thus, no two guests have the same profile. Thus the maximum number of guests possible is 2²⁵