Question 1
How many 4-digit numbers with distinct digits are even?
Step-by-Step Solution:
Concept: Counting even/odd numbers with distinct digits.
Given: 4-digit numbers, distinct digits, even numbers.
Case Analysis for even numbers:
Case 1: Last digit = 0
- First digit: 1-9 (9 choices)
- Remaining 2 digits: choose from remaining 8 digits and arrange
- Ways = 9 × P(8, 2) = 9 × 56 = 504
Case 2: Last digit = 2,4,6,8 (4 choices)
- First digit: cannot be 0 and cannot be last digit (8 choices)
- Remaining 2 digits: choose from remaining 8 digits and arrange
- Ways = 4 × 8 × P(8, 2) = 1792
Total = 504 + 1792 = 2296
Key Principle: Always handle first digit (can't be 0) and last digit (parity constraint) separately.
Concept: Counting even/odd numbers with distinct digits.
Given: 4-digit numbers, distinct digits, even numbers.
Case Analysis for even numbers:
Case 1: Last digit = 0
- First digit: 1-9 (9 choices)
- Remaining 2 digits: choose from remaining 8 digits and arrange
- Ways = 9 × P(8, 2) = 9 × 56 = 504
Case 2: Last digit = 2,4,6,8 (4 choices)
- First digit: cannot be 0 and cannot be last digit (8 choices)
- Remaining 2 digits: choose from remaining 8 digits and arrange
- Ways = 4 × 8 × P(8, 2) = 1792
Total = 504 + 1792 = 2296
Key Principle: Always handle first digit (can't be 0) and last digit (parity constraint) separately.