We are asked to find how many positive 5-digit numbers exist such that their digits, when read from left-to-right, are in weakly decreasing order. Weakly decreasing order means that each digit is either the same or smaller than the one before it.

Step 1: Understanding the problem

Each 5-digit number can be expressed as a sequence of 5 digits $(d_1, d_2, d_3, d_4, d_5)$ where $d_1 \geq d_2 \geq d_3 \geq d_4 \geq d_5$, and each $d_i$ is an integer between 0 and 9. Additionally, since it's a 5-digit number, $d_1 \neq 0$ (the first digit cannot be zero).

Step 2: Mapping the problem to a combinatorial framework

To solve this problem, we can use the idea of stars and bars. The digits must be selected from the set $\{9, 8, 7, 6, 5, 4, 3, 2, 1, 0\}$, with repetition allowed and order restricted to weakly decreasing sequences. The problem is equivalent to finding how many ways we can assign 5 digits to the 10 possible digit values (0 through 9).

Step 3: Handling the restriction $d_1 \neq 0$

If we ignore the restriction that the number is a 5-digit number and consider weakly decreasing sequences of 5 digits (including 0), this becomes a standard stars and bars problem: how many ways can we choose 5 digits from 10, allowing repetition?

This is the same as finding the number of weakly decreasing sequences from $\{9, 8, 7, 6, 5, 4, 3, 2, 1, 0\}$. The number of such sequences is given by the number of ways to place 5 indistinguishable "stars" (the digits) into 10 distinguishable "boxes" (the digit values). This is computed by:

$\binom{10 + 5 - 1}{5} = \binom{14}{5} = 2002$

However, some of these sequences will have the first digit as zero, which does not correspond to a valid 5-digit number.

Step 4: Excluding sequences where $d_1 = 0$

If $d_1 = 0$, the entire sequence must be composed of digits between 0 and 9 that are weakly decreasing. This is equivalent to finding how many ways we can choose 5 digits from $\{0, 0, 0, 0, 0\}$, which is only possible in one way (all digits are zero).

Thus, the number of invalid sequences where $d_1 = 0$ is $1$.

Step 5: Final Calculation

The total number of valid 5-digit numbers is:

$\binom{14}{5} - 1 = 2002 - 1 = 2001$

Thus, there are 2001 positive 5-digit numbers whose digits, when read from left-to-right, are in weakly decreasing order.

Would you like more details on any part of this explanation?

Here are 5 related questions to expand your understanding:

1. How would the answer change if the number of digits was 6 instead of 5?
2. How does the stars and bars method work in combinatorics?
3. What would be the case if we required the digits to be strictly decreasing?
4. How would the solution change if zero was not allowed in any of the digits?
5. Can you derive a formula for the number of n-digit numbers whose digits are in weakly decreasing order?

Tip: In combinatorics, "stars and bars" is a powerful method to handle problems of distributing indistinguishable objects into distinguishable boxes. It is especially useful for finding combinations with repetition.