求助！！！此题求高手讲解~~

sukigmat

A three-digit code for certain logs uses the digits 0,1,2,3,4,5,6,7,8,9 according to the following constraints. The first digit cannot be 0 or 1, the second digit must be 0 or 1, and the second and third digits cannot both be 0 in the same code. How many different codes are possible?
A.144 B.152 C.160 D.168 E.176
哪位牛人知道怎么算哒？
我算出来是152，可答案是144~~
第一位2~9是8种可能，如果第二位取0，有9种可能；如果第二位取1，有十种可能，所以8*9+8*10=152，哪里错了？

sukigmat

是答案错了么？？