Problem:
Find the number of 3-digit numbers such that the digit 1 is never to the immediate right of 2.

Solution:
To solve this problem, we can consider the possible positions of the digits in a 3-digit number and count the number of valid arrangements.

Case 1: 1 is in the first position:
If 1 is in the first position, the second and third digits can be any digit from 0 to 9 except 2. Thus, there are 9 choices for the second digit and 10 choices for the third digit. So, the total number of valid arrangements in this case is 9 * 10 = 90.

Case 2: 1 is in the second position:
If 1 is in the second position, the first digit can be any digit from 1 to 9 except 2. The third digit can be any digit from 0 to 9 except 1. Thus, there are 8 choices for the first digit and 9 choices for the third digit. So, the total number of valid arrangements in this case is 8 * 9 = 72.

Case 3: 1 is in the third position:
If 1 is in the third position, the first and second digits can be any digit from 1 to 9 except 2. Thus, there are 8 choices for the first digit and 9 choices for the second digit. So, the total number of valid arrangements in this case is 8 * 9 = 72.

Total number of valid arrangements:
To find the total number of valid arrangements, we sum up the number of valid arrangements from each case:
Total = 90 + 72 + 72 = 234

However, we need to exclude the numbers with leading zeroes, as they are not 3-digit numbers.
There are 9 choices for the first digit (excluding 0) and 9 choices for the second digit (excluding 1). The third digit can be any digit from 0 to 9 except 1. Thus, there are 9 * 9 * 9 = 729 valid 3-digit numbers.

Final answer:
The number of 3-digit numbers such that the digit 1 is never to the immediate right of 2 is 729.

Explanation:
The correct answer is option 'C' (881) because it represents the total count of valid 3-digit numbers.