We have the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. We want to form a four-digit number with distinct digits. a) How many such numbers can be formed? b) How many four-digit numbers can be formed, knowing that the number is even? c) How many four-digit numbers can be formed, knowing that the number is odd? d) How many four-digit numbers can be formed, knowing that the number is a multiple of 5?

1. Problem Description

We have the digits 0, 1, 2, 3, 4, 5, 6, 7, 8,

9. We want to form a four-digit number with distinct digits.

a) How many such numbers can be formed?

b) How many four-digit numbers can be formed, knowing that the number is even?

c) How many four-digit numbers can be formed, knowing that the number is odd?

d) How many four-digit numbers can be formed, knowing that the number is a multiple of 5?

2. Solution Steps

a) To form a four-digit number with distinct digits, the first digit cannot be

0. So there are 9 choices for the first digit. After choosing the first digit, we have 9 remaining choices for the second digit (including 0). Then we have 8 choices for the third digit and 7 choices for the fourth digit.

Therefore, the 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. We consider two cases:

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. The number of such numbers is $9 \times 8 \times 7 \times 1 = 504$.

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

8. There are 4 choices for the last digit. The first digit cannot be 0 or the digit chosen for the last digit, so there are 8 choices for the first digit. For the second digit, we can use 0, but not the first digit or the last digit, so there are 8 choices. For the third digit, there are 7 remaining choices. The number of such numbers is $8 \times 8 \times 7 \times 4 = 1792$.

Therefore, 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. There are 5 choices for the last digit. The first digit cannot be 0 or the last digit, so there are 8 choices. For the second digit, we can use 0, but not the first digit or the last digit, so there are 8 choices. For the third digit, there are 7 remaining choices. Therefore, the total number of odd numbers is $8 \times 8 \times 7 \times 5 = 2240$.

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

5. We consider two cases:

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. The number of such numbers is $9 \times 8 \times 7 \times 1 = 504$.

Case 2: The last digit is

5. The first digit cannot be 0 or 5, so there are 8 choices for the first digit. For the second digit, we can use 0, but not the first digit or 5, so there are 8 choices. For the third digit, there are 7 remaining choices. The number of such numbers is $8 \times 8 \times 7 \times 1 = 448$.

Therefore, the total number of multiples of 5 is $504 + 448 = 952$.

3. Final Answer

a) 4536

b) 2296

c) 2240

d) 952