Andrew has a four-digit number whose last digit is $2$. Given that this number is divisible by $9$, determine the number of possible values for this number that Andrew could have. Note that leading zeros are not allowed.
To determine the number of possible four-digit numbers ending in 2 that are divisible by 9, we need to use the property that a number is divisible by 9 if and only if the sum of its digits is divisible by 9.

Let the four-digit number be represented as \( A_1A_2A_3 2 \). The sum of the digits of this number is \( A_1 + A_2 + A_3 + 2 \). For the number to be divisible by 9, this sum must be a multiple of 9.

Thus, we have:
\[ A_1 + A_2 + A_3 + 2 \equiv 0 \pmod{9} \]

This implies:
\[ A_1 + A_2 + A_3 \equiv -2 \pmod{9} \]
\[ A_1 + A_2 + A_3 \equiv 7 \pmod{9} \]

Since \( A_1, A_2, \) and \( A_3 \) are digits (0-9), the possible values for \( A_1 + A_2 + A_3 \) are 7, 16, or 25 (as the maximum sum of three digits is 27).

1. **Case 1: \( A_1 + A_2 + A_3 = 7 \)**

   We need to find the number of non-negative integer solutions to the equation:
   \[ A_1 + A_2 + A_3 = 7 \]
   Using the stars and bars method, the number of solutions is:
   \[ \binom{7 + 2}{2} = \binom{9}{2} = 36 \]

   However, since \( A_1 \) cannot be 0 (as it is a four-digit number), we need to subtract the cases where \( A_1 = 0 \):
   \[ A_2 + A_3 = 7 \]
   The number of solutions to this equation is:
   \[ \binom{7 + 1}{1} = 8 \]

   Therefore, the number of valid solutions for \( A_1 + A_2 + A_3 = 7 \) is:
   \[ 36 - 8 = 28 \]

2. **Case 2: \( A_1 + A_2 + A_3 = 16 \)**

   We need to find the number of non-negative integer solutions to the equation:
   \[ A_1 + A_2 + A_3 = 16 \]
   Using the stars and bars method, the number of solutions is:
   \[ \binom{16 + 2}{2} = \binom{18}{2} = 153 \]

   Again, since \( A_1 \) cannot be 0, we need to subtract the cases where \( A_1 = 0 \):
   \[ A_2 + A_3 = 16 \]
   The number of solutions to this equation is:
   \[ \binom{16 + 1}{1} = 17 \]

   Therefore, the number of valid solutions for \( A_1 + A_2 + A_3 = 16 \) is:
   \[ 153 - 17 = 136 \]

3. **Case 3: \( A_1 + A_2 + A_3 = 25 \)**

   We need to find the number of non-negative integer solutions to the equation:
   \[ A_1 + A_2 + A_3 = 25 \]
   Using the stars and bars method, the number of solutions is:
   \[ \binom{25 + 2}{2} = \binom{27}{2} = 351 \]

   Again, since \( A_1 \) cannot be 0, we need to subtract the cases where \( A_1 = 0 \):
   \[ A_2 + A_3 = 25 \]
   The number of solutions to this equation is:
   \[ \binom{25 + 1}{1} = 26 \]

   Therefore, the number of valid solutions for \( A_1 + A_2 + A_3 = 25 \) is:
   \[ 351 - 26 = 325 \]

Summing up all the valid solutions, we get:
\[ 28 + 136 + 325 = 489 \]

The final answer is \(\boxed{489}\)