How many 4-digit numbers have non-increasing digits?

To find how many 4-digit numbers have non-increasing digits, we need to count all possible 4-digit numbers where each digit is greater than or equal to the next digit. Let's represent a 4-digit number as \(abcd\) where \(a \geq b \geq c \geq d\) and each of \(a\), \(b\), \(c\), and \(d\) are digits from 0 to 9.

### Solution Outline

1. **Define the Range for Each Digit**: Since we are dealing with 4-digit numbers, \(a\) (the thousands place) must be non-zero (i.e., \(a \geq 1\)), while \(b\), \(c\), and \(d\) can each be any digit from 0 to 9.

2. **Apply Combinatorics**: We need to find combinations of digits such that they are non-increasing. To do this, we can use a combinatorial approach with the concept of "stars and bars" or "combinations with repetition."

3. **Transform the Problem**: Let's define a new variable to count how often each digit appears in our number. For instance, let:
   - \(x_1\) represent the count of 9s,
   - \(x_2\) the count of 8s,
   - \(x_3\) the count of 7s,
   - ...
   - \(x_{10}\) represent the count of 0s.

   These counts satisfy:
   \[
   x_1 + x_2 + \cdots + x_{10} = 4
   \]
   because the total number of digits in any 4-digit number is 4.

4. **Constraints**: Ensure that the first non-zero digit appears in the thousands place to avoid 0 starting the number. We can handle this by subtracting combinations that start with zero.

5. **Calculate Using Combinations with Repetition**:
   The number of solutions to the equation \(x_1 + x_2 + \cdots + x_{10} = 4\) (where each \(x_i \geq 0\)) is given by the formula for combinations with repetition:
   \[
   \binom{4 + 10 - 1}{10 - 1} = \binom{13}{9} = 715
   \]

After accounting for restrictions on the first digit (ensuring it’s non-zero), this approach provides the count of all valid non-increasing 4-digit numbers.

### Final Answer
There are **715** four-digit numbers with non-increasing digits.

Would you like more details on the steps? 

1. How do we handle non-increasing sequences in other lengths, like 5-digit numbers?
2. Can we use similar combinatorial approaches for non-decreasing digits?
3. How do we calculate combinations with constraints on specific digits?
4. Could we explore examples of 4-digit numbers with repeating digits?
5. What’s the role of “stars and bars” in combinatorial math?

**Tip:** Non-increasing sequences follow a systematic pattern, making them ideal for combinatorics-based counting.

Explore how to count 4-digit numbers with non-increasing digits using combinatorial methods. This solution covers key concepts in combinatorics, including the 'Stars and Bars' theorem and combinations with repetition, making it ideal for students in Grades 11-12.

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