UPSC NDA II 2015 Mathematics Paper

Concept:Counting 3-digit even numbers using permutations without repetition, with special handling for zero at different positions.Explanation:We have digits $0,1,2,3,4,5$. We need 3-digit even numbers (units digit even) without repeating digits. We consider three cases based on the position of zero.Case 1: Units digit = $0$ (fixed). Then tens digit can be any of the remaining $5$ digits ($1,2,3,4,5$). Hundreds digit can be any of the remaining $4$ digits (excluding the two used, and cannot be $0$). Number of such numbers = $1 \times 5 \times 4 = 20$.Case 2: Tens digit = $0$, and units digit is $2$ or $4$. Units digit: $2$ choices. Tens digit: fixed as $0$ ($1$ choice). Hundreds digit: any of the remaining $4$ digits (cannot be $0$, excludes the units digit). Number = $2 \times 1 \times 4 = 8$.Case 3: No zero in the number. Units digit must be $2$ or $4$: $2$ choices. Tens digit: any of the remaining $4$ digits (excluding the units digit and $0$). Hundreds digit: any of the remaining $3$ digits. Number = $2 \times 4 \times 3 = 24$.Total = $20 + 8 + 24 = 52$.Answer:$52$ (Option C)