PUMaC 2010 · 组合（A 组） · 第 2 题

PUMaC 2010 — Combinatorics (Division A) — Problem 2

Let xyz represent the three-digit number with hundreds digit x , tens digit y , and units digit z , and similarly let yz represent the two-digit number with tens digit y and units digit z . How many three-digit numbers abc , none of whose digits are 0, are there such that ab > bc > ca ?

解析

Let xyz represent the three-digit number with hundreds digit x , tens digit y , and units digit z , and similarly let yz represent the two-digit number with tens digit y and units digit z . How many three-digit numbers abc , none of whose digits are 0, are there such that ab > bc > ca ? Answer: 120 Solution: If one two-digit number is greater than another, then the tens digit of the first num- ber must be greater than or equal to the tens digit of the second number. Therefore, if abc satisfies the given condition, then a ≥ b ≥ c . Now note that if b = c , then since c ≤ a , we have bc ≤ ca , a contradiction. Therefore, a ≥ b > c . Conversely, if a ≥ b > c , then we can easily see that abc satisfies the given condition. Therefore, the problem is equivalent to finding the number of ordered pairs ( a, b, c ) of integers between 1 and 9 such that a ≥ b > c . If a = b > c , we can produce all such ( a, b, c ) by choosing 2 of the integers between 1 and 9, and setting a and b equal to the larger integer, and c equal to the smaller integer. If a > b > c , we can produce all such ( a, b, c ) by choosing 3 of the integers between 1 and 9, and setting a equal to the largest integer, b equal to the middle integer, and c equal to the smallest integer. ( ) ( ) 9 9 Therefore the answer is + = 45 + 120 = 120 . 2 3