NDA Mathematics Question (2026) — Solution

Step-by-step solution

Total number of 4-digit numbers formed using 0, 1, 2, 3, 4 without repetition: The thousands place can be filled in 4 ways (excluding 0). The remaining 3 places can be filled in ^ 4 P_ 3 = 24 ways. Total possible numbers = 4 24 = 96. A number is divisible by 4 if the number formed by its last two digits is divisible by 4. The possible last two digits from the given set are 04, 12, 20, 24, 32, 40. Case 1: Last two digits are 04, 20, or 40 (3 cases). Since 0 is used, the first two places can be filled from the remaining 3 digits in ^ 3 P_ 2 = 6 ways. Number of such numbers = 3 6 = 18. Case 2: Last two digits are 12, 24, or 32 (3 cases). Since 0 is not used in the last two digits, the thousands place cannot be 0. It can be filled in 2 ways. The hundreds place can be filled in 2 ways. Number of such numbers = 3 (2 2) = 12. Total numbers divisible by 4 = 18 + 12 = 30. Required probability = 30 96 = 5 16 . Answer: 5 16