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 3 if the sum of its digits is a multiple of 3. The sum of all given digits is 0 + 1 + 2 + 3 + 4 = 10. To choose 4 digits whose sum is a multiple of 3, the excluded digit must leave a remainder of 1 when divided by 3. Thus, the excluded digit can be 1 or 4. Case 1: Excluding 1, the digits are 0, 2, 3, 4. Number of 4-digit numbers = 3 3! = 18. Case 2: Excluding 4, the digits are 0, 1, 2, 3. Number of 4-digit numbers = 3 3! = 18. Total favorable outcomes = 18 + 18 = 36. Required probability = 36 96 = 3 8 . Answer: 3 8