HMMT 二月 2026 · COMB 赛 · 第 3 题

HMMT February 2026 — COMB Round — Problem 3

The numbers 1 , 2 , 3 , 4 , 5 , 6 , and 7 are written on a blackboard in some order. Jacob repeatedly swaps numbers at adjacent positions on the blackboard until the numbers are sorted in ascending order. Compute the number of initial orderings for which it is possible that the number 4 was included in a swap at most once.

解析

The numbers 1 , 2 , 3 , 4 , 5 , 6 , and 7 are written on a blackboard in some order. Jacob repeatedly swaps numbers at adjacent positions on the blackboard until the numbers are sorted in ascending order. ©2026 HMMT Compute the number of initial orderings for which it is possible that the number 4 was included in a swap at most once. Proposed by: Sebastian Attlan Answer: 324 Solution: We split into cases based on whether 4 is part of 0 or 1 swap. • If 4 is part of no swaps , then 1 , 2 , and 3 all start on the left of 4 , while 5 , 6 , and 7 all start on the right of 4 . Regardless of the initial ordering of the 1 , 2 , and 3 , it is possible to rearrange them into the correct order without performing swaps involving 4 . Similarly, the initial ordering of the 4 , 5 , and 6 is unconstrained. There are 3! ways to order the numbers on each side of 4 , so 2 the number of such permutations is 3! = 36 . • If 4 is part of one swap , there are 6 ways to pick the number it needs to swap with. Regardless of which number is selected, this fixes four numbers on one side of 4 and two numbers on the other. By an analogous argument to the first case, the ordering within each side is unconstrained. Thus, the number of such permutations is 6 · 4! · 2! = 288 . Thus, the answer is 36 + 288 = 324 .