Euler problem 24.1.1

Posted on by

To formalize, if a_0 < \ldots < a_n , then in the k-th permutation of \{a_0, \ldots, a_n\} in lexicographic order, the leading entry is a_q if k = q(n!) + r for some q \geq 0 and 0 < r \leq n! . (Note that the definition of r here is a bit different from the usual remainder, for which 0 \leq r < n! . Also, a_q is the (q+1)-th entry but not the q-th entry in the sequence, because the index starts from 0.)

— edited Aug 30, 2011 at 18:23
— answered Aug 30, 2011 at 17:59
— user1551
— math stackexchange

Why r Cannot Be Zero: If r were allowed to be zero, it would imply that the leading entry of the permutation could be determined without any remaining elements to choose from, which contradicts the requirement of having a valid permutation. In other words, a remainder of zero would mean that we have perfectly divided k by n! , leading us to a situation where we would not be able to select the next element in the permutation sequence.

— AI Assistant

.

.

2024.10.30 Wednesday (c) All rights reserved by ACHK