UPSC NDA I 2026 Mathematics Paper

Concept:Total 4-digit numbers without repetition, then count those whose last two digits are divisible by 4.Explanation:Total 4-digit numbers formed from $0,1,2,3,4$ without repetition:First digit cannot be $0$, so total $= 4 \times 4 \times 3 \times 2 = 96$.For divisibility by $4$, the last two digits must be divisible by $4$.Possible valid endings from the digits are:$04, 12, 20, 24, 32, 40$.Endings containing $0$: $04, 20, 40$.For each, $0$ is already used, so the first two digits can be arranged in $3 \times 2 = 6$ ways.Contribution $= 3 \times 6 = 18$.Endings not containing $0$: $12, 24, 32$.For each, $0$ remains available, so the first digit has $2$ choices and the second has $2$ choices.Contribution $= 3 \times 4 = 12$.Total favourable numbers $= 18 + 12 = 30$.Required probability $= \frac{30}{96} = \frac{5}{16}$.Answer:$\frac{5}{16}$