Without repetition, a five -digit number can be formed using the five digits in 5! ways (5 × 4 × 3 × 2 × 1) Out of these 5! numbers, 4! numbers will be starting with digit 0. (0 (fixed) 4 × 3 × 2 × 1) ∴ Total number of 5–digit numbers that can be formed using the digits 0, 1, 2, 3, 4 = 5! – 4! = 120 – 24 = 96. ⇒ No. of exhaustive cases = n(S) = 96 Now, only these numbers are divisible by 4 in which the numbers formed by the last two digits is divisible by 4. Thus, the numbers ending in 04, 12, 20, 24, 32 and 40 will be divisible by 4. (i) If the numbers end in 04, the remaining three numbers 1, 2, 3, can be arranged in 3! ways = 6 ways (ii) If the numbers end in 20, the remaining three numbers, i.e., 1, 4, 3 can be arranged in 3! ways = 6 ways (iii) If the numbers end in 40, the remaining three numbers, i.e., 1, 2, 3 can be arranged in 3! ways = 6 ways (iv) If the numbers end in 12, the remaining three numbers, i.e., 0, 3, 4 can be arranged 3! ways, but these cases in which 0 is the extreme left digit one to be discorded. The number of such cases is 2!. ∴ Number of numbers ending in 12 = 3! – 2! = 4 (v) Similarly the numbers ending in 24 and 32 are 4 each. ∴ Total number of favourable cases, i.e., number of 5–digit numbers divisible by 4 = 6 + 6 + 6 + 4 + 4 + 4 = 30∴ Required probability = \(rac{30}{96}\) = \(rac{5}{16}.\)