We are given the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. We want to form a 4-digit number using these digits, where all the digits must be different. a) How many such numbers can be formed? b) How many of these numbers are even? c) How many of these numbers are odd? d) How many of these numbers are multiples of 5?

1. Problem Description

We are given the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and

9. We want to form a 4-digit number using these digits, where all the digits must be different.

a) How many such numbers can be formed?

b) How many of these numbers are even?

c) How many of these numbers are odd?

d) How many of these numbers are multiples of 5?

2. Solution Steps

a) We want to find the number of 4-digit numbers with distinct digits chosen from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Since the first digit cannot be 0, there are 9 choices for the first digit. Once the first digit is chosen, there are 9 choices for the second digit (since we can now use 0). Then there are 8 choices for the third digit, and 7 choices for the fourth digit.

Therefore, the total number of such numbers is $9 \times 9 \times 8 \times 7$.

$9 \times 9 \times 8 \times 7 = 81 \times 56 = 4536$

b) For the number to be even, the last digit must be 0, 2, 4, 6, or

8.

Case 1: The last digit is

0. Then there are 9 choices for the first digit, 8 choices for the second digit, and 7 choices for the third digit. So there are $9 \times 8 \times 7 = 504$ such numbers.

Case 2: The last digit is 2, 4, 6, or

8. There are 4 choices for the last digit. Since the first digit cannot be 0 or the last digit, there are 8 choices for the first digit. Then there are 8 choices for the second digit (since 0 can now be used, but the last two chosen digits cannot), and 7 choices for the third digit. So there are $8 \times 8 \times 7 \times 4 = 1792$ such numbers.

Thus, the total number of even numbers is $504 + 1792 = 2296$.

c) For the number to be odd, the last digit must be 1, 3, 5, 7, or

9. So there are 5 choices for the last digit. The first digit cannot be 0 or the last digit, so we have two subcases.

Case 1: The first digit is

0. Then there are 8 choices for the first digit.

If the first digit is zero, this is impossible since a four digit number cannot start with zero.

Case 2: The first digit is nonzero. The first digit cannot be the last digit, nor zero. So there are 8 possibilities. The second digit cannot be the first digit, nor the last digit, so there are 8 possibilities. The third digit cannot be the first, second nor last digits, so there are 7 possibilities. The last digit has five possibilities.

Therefore the possibilities are $8 \times 8 \times 7 \times 5 = 2240$

d) For the number to be a multiple of 5, the last digit must be 0 or

5.

Case 1: The last digit is

0. Then there are 9 choices for the first digit, 8 choices for the second digit, and 7 choices for the third digit. So there are $9 \times 8 \times 7 = 504$ such numbers.

Case 2: The last digit is

5. The first digit cannot be 0 or 5, so there are 8 choices for the first digit. The second digit can be 0 but not 5 and not the first digit, so there are 8 choices. The third digit cannot be the first, second, or last digits, so there are 7 choices. Thus, there are $8 \times 8 \times 7 = 448$ such numbers.

The total number of multiples of 5 is $504 + 448 = 952$.

3. Final Answer

a) 4536

b) 2296

c) 2240

d) 952