If all distinct permutations of the letters in “MASTER” are arranged in…

Concept

In lexicographic order, permutations are grouped by their earliest differing letter.

For n distinct letters, fixing the first k positions leaves (n − k)! arrangements. These block sizes locate a rank without listing every permutation.

Application

1. Arrange the letters alphabetically: A, E, M, R, S, T.

2. Fixing the first letter leaves 5! = 120 arrangements. Therefore ranks 1–120 begin with A, so the 49th word begins with A.

3. After A, each choice of the second letter creates 4! = 24 arrangements.

4. The AE block covers ranks 1–24, and the AM block covers ranks 25–48. Therefore rank 49 is the first word in the AR block.

5. For the first word in the AR block, place the remaining letters E, M, S, T in alphabetical order. This gives AREMST.

Cross-check

Exactly 24 + 24 = 48 words precede the AR block. Hence its first member has rank 48 + 1 = 49, confirming AREMST.