Forming Four-Digit Numbers with Specific Restrictions

In this article, we will explore the problem of forming four-digit numbers using the digits from 0 to 9, with a specific restriction: the last digit must be zero, and repetitions are not allowed. We will work through the steps to determine how many such numbers can be formed, using permutations and logical reasoning.

Understanding the Problem

Given a set of ten digits (0-9), we need to form four-digit numbers where:

The last digit must be 0. No digit can be repeated. The first digit cannot be 0, as it would not be a four-digit number.

Step-by-Step Solution

Fixing the Last Digit

Since the last digit must be 0, we have:

(text{Last digit} 0)

Choosing the First Three Digits

We need to choose the first, second, and third digits from the remaining set of nine digits (1-9), with no repetition allowed.

Calculating the Choices

First digit (W): 9 choices, since it cannot be 0. Second digit (X): 8 choices, as one digit has already been used for the first position. Third digit (Y): 7 choices, after the first two digits have been used.

Total Combinations

The total number of ways to form the first three digits is calculated as:

(9 times 8 times 7 504)

Conclusion

Therefore, the total number of four-digit numbers that can be formed under the specified conditions is 504.

Alternatively, if we represent the four-digit number as ABCD, where D (the last digit) is 0, we have:

A (the first digit): 9 choices (1-9). B and C (middle digits): 8 and 7 choices respectively, after selecting A.

So the total number of such numbers is:

(9 times 8 times 8 times 7 4032)

Explanation of Different Methods

Let's explore another way to approach this problem. We can write a four-digit number as ABCD, where:

Decimal value: (N 1000A 100B 10C D) A (the first digit): 9 choices (1-9). B and C (second and third digits): 8 and 7 choices respectively, after selecting A and considering no repetition. D (the last digit): 1 choice (0).

Thus, the total number of such numbers is:

(9 times 8 times 8 times 7 4032)

Generalization

In general, when forming a k-digit number with the last digit fixed as 0 and no repetitions:

Total choices for the first k-1 digits: ((10 - 1) times (10 - 2) times ... times (10 - (k-1))) Total numbers: (9 times 8 times 8 times 7 4032)

This method can be applied to form four-digit numbers with the last digit fixed as 0 and no repetitions, leading to a total of 504 combinations.