請教一題OG MATH

Of the three-digit integers greater than 700,how many have two digit that are equal to each other and the remaining digit different from the other two?

答案是:80

請問能教我解法嗎?

一个苯苯的方法:

首先这个数的范围是700<X<=999

(1)711,722,733,744,755,766,.....799 共有8个数(没有777)

(2)800,811,......899 共有9 个数(没有888)

(3)900,911,922........988 共有9 个数(没有999)

(4)770,771,772,.......779 (除去777) 共有8个数

(5)880-889 (除去888)共有9个

(6)990-998 (除去999) 共有9个

(7)707,717,727,737........797(除去777) 共有9个,

(8)808,818,828,,,,,,,,,898 (除去888)共有9个

(9)909,919,929,,,,,,,,,989 (除去999)共有9个,

所以一共是: 8*9+8=80