A 4-digit falling number is a number where each digit is greater than the digit to its right. For example, 4321 is a 4-digit falling number, but 4322 is not.

To form a 4-digit falling number, we must select 4 distinct digits from the 10 decimal digits (0 through 9) and arrange them in strictly decreasing order. Since there are no repeats and the digits must be in a strict order, each selection of 4 digits automatically corresponds to exactly one falling number.

Steps:

1. Choose 4 digits from the 10 available digits: The number of ways to choose 4 distinct digits from 10 is calculated using combinations: $\binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$

2. Arranging the digits: Since the digits must be arranged in strictly decreasing order, there is only one way to arrange any set of 4 digits.

Thus, the total number of 4-digit falling numbers is equal to the number of ways to choose 4 distinct digits from the 10 available digits, which is $\binom{10}{4} = 210$.

Final Answer:

The number of 4-digit falling numbers is $\boxed{210}$.

Would you like further details or have any questions?

Here are 5 related questions:

1. How many 3-digit falling numbers can be formed?
2. What if the digits could repeat in falling numbers? How would the count change?
3. How many 5-digit falling numbers can we create from decimal digits?
4. How does this problem change for rising numbers, where each digit is smaller than the next?
5. Can you calculate the number of 4-digit falling numbers without using the digit 0?

Tip: Remember that falling numbers are strictly decreasing sequences, so once you've selected the digits, the arrangement is automatic!