Total no. of 4–digit numbers (Repetition of digits not allowed ) : I can be filled in 4 ways as it can’t have zero, II can be filled in 4 ways by the remaining 4 digits, III in 3 ways and IV in 2 ways ∴ Total no. of outcomes =4×4×3×2=96 Similarly, No. of 4-digit numbers ending with 0 =4×3×2=24 No. of 4-digit numbers ending with 2=3×3×2=18 No. of 4-digit numbers ending with 4=3×3×2=18 No. of favourable outcomes =24+18+18=60 Required probability =