The problem asks how many tickets exist in the range 0000 to 9999, where the sum of the digits equals 21.

Let's define the four-digit number as $a_1a_2a_3a_4$, where $a_1, a_2, a_3, a_4$ represent the digits, each of which ranges from 0 to 9. The condition is that the sum of the digits equals 21, i.e.,

$a_1 + a_2 + a_3 + a_4 = 21$

To solve this, we need to determine how many such combinations exist for $a_1, a_2, a_3, a_4$, where each $a_i$ is a digit (between 0 and 9).

Step-by-step approach:

1. Total possible sum: The sum of four digits can be at most 36 (if all digits are 9, i.e., $9+9+9+9 = 36$).

2. Restriction: The sum is specifically 21.

3. Non-negative integer solution: The problem of finding how many non-negative integer solutions there are to the equation $a_1 + a_2 + a_3 + a_4 = 21$, where each $a_i \leq 9$, can be solved using a combinatorial approach.

4. Transformation: We transform this into a stars-and-bars problem with restrictions.

I'll calculate this next.There are 592 tickets in the range from 0000 to 9999 whose digits add up to 21.

Would you like a detailed explanation of how this was calculated? Here are some related questions for further exploration:

1. How would the solution change if the sum of the digits needed to be 20 instead of 21?
2. What is the total number of possible tickets between 0000 and 9999?
3. How can this problem be extended to tickets with more than four digits?
4. How would we approach this if the digits were restricted to a different range, such as 1 to 5?
5. What combinatorial techniques are useful for solving similar problems with digit constraints?

Tip: Problems like these, where sums of digits are constrained, often involve combinatorics and stars-and-bars techniques with additional restrictions.