When all the words in "MASTER" is sorted in alphabetical order then what is…

To find the k-th word when all arrangements of a word's letters (all distinct) are listed in dictionary (alphabetical) order, first sort the letters alphabetically. Fix the first letter and count how many arrangements begin with it — this count equals (n - 1)! where n is the number of letters not yet fixed. Keep adding each candidate letter's block, in alphabetical order, until the running total reaches or passes the required rank; the word lies in that block. Repeat the same counting logic for the next position, using only the letters not yet placed.

1. Arrange the six distinct letters of MASTER alphabetically: A, E, M, R, S, T.

2. Words beginning with A: the remaining 5 letters can be arranged in 5! = 120 ways, so these occupy ranks 1 to 120 — rank 49 lies inside this block, so the word begins with A.

3. Within the 'A' block, fix the second letter from the remaining set {E, M, R, S, T} in alphabetical order. Words starting 'AE' occupy 4! = 24 ranks (1–24 overall) and words starting 'AM' occupy the next 24 ranks (25–48 overall) — together these two sub-blocks account for 48 words.

4. Since 48 words have already been counted, rank 49 is the first word of the next sub-block, which begins with the pair 'A' followed by 'R' (the next available letter after 'M' in the alphabetical list).

5. The remaining four letters — E, M, S, T — must appear in strictly increasing alphabetical order to form the smallest (first) word of this sub-block: E, M, S, T.

6. Putting it together, the 49th word is A, R, E, M, S, T, i.e., AREMST.

Verification: 24 (AE) + 24 (AM) = 48 words precede this sub-block, so the 49th word is exactly the first word beginning 'AR'; and E, M, S, T is indeed the smallest ordering of the four leftover letters, confirming AREMST as the 49th word.