How many different 4-digit numbers greater than 3000 can be formed using the digits 2, 3, 4, 5 and 6, without repetition, such that the number formed is even?

Digits: 2, 3, 4, 5, 6 Number must be 4-digit, even, and > 3000. Last digit (units) must be even: 2, 4, or 6. Case 1: Units = 2 Thousands digit: can be 3, 4, 5, 6 (4 options). Remaining 3 digits fill 2 middle places: 3P2 = 6 ways. Total = 4 × 6 = 24 Case 2: Units = 4 Thousands digit: from {2, 3, 5, 6}, but must be > 3 → 3, 5, 6 (3 options). Middle two digits: remaining 3 digits → 3P2 = 6 ways. Total = 3 × 6 = 18 Case 3: Units = 6 Thousands digit: from {2, 3, 4, 5}, but must be > 3 → 4, 5, 3 (3 options). Middle two digits: remaining 3 digits → 3P2 = 6 ways. Total = 3 × 6 = 18 Overall total = 24 + 18 + 18 = 60