1. Problem: Find the number of different three-digit numbers under various conditions. 2. Formula and rules: For counting numbers with digits, use permutations and combinations. 3. a) Three-digit numbers where digits can repeat: - The hundreds digit can be from 1 to 9 (9 options, since 0 cannot be the first digit). - The tens digit can be from 0 to 9 (10 options). - The units digit can be from 0 to 9 (10 options). - Total numbers = $9 \times 10 \times 10 = 900$. 4. b) Three-digit numbers where digits cannot repeat: - Hundreds digit: 9 options (1-9). - Tens digit: 9 options (0-9 except the hundreds digit). - Units digit: 8 options (0-9 except hundreds and tens digits). - Total numbers = $9 \times 9 \times 8 = 648$. 5. c) Three-digit numbers where the hundreds digit is a prime number and digits can repeat: - Prime digits for hundreds place: 2, 3, 5, 7 (4 options). - Tens digit: 10 options (0-9). - Units digit: 10 options (0-9). - Total numbers = $4 \times 10 \times 10 = 400$. 6. d) Three-digit numbers where the hundreds digit is even and digits cannot repeat: - Even digits for hundreds place: 2, 4, 6, 8 (4 options). - Tens digit: 9 options (0-9 except hundreds digit). - Units digit: 8 options (0-9 except hundreds and tens digits). - Total numbers = $4 \times 9 \times 8 = 288$. Final answers: - a) 900 - b) 648 - c) 400 - d) 288