How many integers between 10 and 999 have the sum of their digits equal to 9?
Solution 1: Let's divide it into two cases: two-digit numbers and three-digit numbers. In the case of two-digit numbers, we have $18, 27, 36, 45, 54, 63$, 72, 81, and 90, for a total of 9 numbers. Similarly, we list the three-digit numbers as follows:

$$
\begin{aligned}
108,117,126,135,144,153,162,171,180 & \leadsto 9 \text { numbers; } \\
207,216,225,234,243,252,261,270 & \leadsto 8 \text { numbers; } \\
306,315,324,333,342,351,360 & \leadsto 7 \text { numbers; } \\
405,414,423,432,441,450 & \leadsto 6 \text { numbers; } \\
504,513,522,531,540 & \leadsto 5 \text { numbers; } \\
603,612,621,630 & \leadsto 4 \text { numbers; } \\
702,711,720 & \leadsto 3 \text { numbers; } \\
801,810 & \leadsto 2 \text { numbers; } \\
900 & \leadsto 1 \text { number. }
\end{aligned}
$$

Therefore, we have $9+8+7+6+5+4+3+2+1=45$ three-digit numbers. Thus, there are $9+45=54$ numbers between 10 and 999 whose sum of digits is equal to 9.

Solution 2: In the case of two-digit numbers, once the tens digit is chosen, the units digit is automatically defined. Since the tens digit can be $1,2,3,4,5,6,7,8$ or 9, we have nine two-digit numbers such that the sum of their digits is 9.

In the case of three-digit numbers, denoting by $n$ the hundreds digit, the sum of the tens and units digits is $9-n$, so there are $9-n+1=10-n$ possible choices for the tens digit, which can be 0, and the units digit is automatically defined. Since the hundreds digit can be $1,2,3,4,5,6,7,8$ or 9, we have

$(10-1)+(10-2)+(10-3)+(10-4)+(10-5)+(10-6)+(10-7)+(10-8)+(10-9)=45$

three-digit numbers such that the sum of their digits is 9. Thus, there are $9+45=54$ numbers between 10 and 999 whose sum of digits is equal to 9.