1. **Problem Statement:** Find the shortest and longest distances between permutations of the letters in "ritangle" (8 distinct letters) considering all $8!$ arrangements. 2. Since all letters are distinct, the total permutations are $8! = 40320$. 3. The problem implies using the permutations as points in some metric space. Typically, the distance between two permutations can be measured by the number of differing positions or related metrics. However, the problem does not explicitly define the distance measure. Assuming standard Hamming distance between permutations where each permutation is a sequence of 8 letters: 4. The shortest distance between any two distinct permutations is 2, because swapping just two letters changes exactly 2 positions. 5. The longest distance (maximal number of differing positions) is 8, when two permutations have completely different letters at every position. 6. Thus, among all $8!$ permutations: - Shortest distance between any two distinct permutations is 2. - Longest distance between two permutations is 8. This follows from the combinatorial properties of permutations. **Final answer:** Shortest distance = $2$, Longest distance = $8$.