11) Fabio finds an old combination lock; to open the lock, one must align three digits in the correct order, each of which can vary from 0 to 9. Fabio does not remember the correct combination, but he is sure that the sum of the three digits is 10. How many attempts will he have to make, at most, to find the correct combination? $\begin{array}{llll}\text { (A) } 61, & \text { (B) } 63, \text { (C) } 65, & \text { (D) } 67, & \text { (E) } 69 .\end{array}$
11. The answer is (B). Let $X, Y, Z$ be the first, second, and third digit of the lock. To solve the problem, we need to count all possible ordered triples $(X, Y, Z)$ such that $X+Y+Z=10$. If $X=0$, then $Y+Z=10$; in this case, neither $Y$ nor $Z$ can be $0$, $Y$ can take all values between 1 and 9, inclusive, and for each choice of $Y$, the value of $Z$ is determined: $Z=10-X$; there are thus 9 triples with $X=0$. If $X=1$, then $Y+Z=9$, $Y$ can take all values between 0 and 9, inclusive, and $Z$ is consequently determined; there are then 10 triples with $X=1$. We can proceed in this way finding that there are 9 triples with $X=2$, 8 with $X=3$, and so on until 2 triples with $X=9$. The total number of triples is then $9+10+9+8+7+6+5+4+3+2=63$.