UPSC NDA I 2017 Mathematics Solved Paper

Concept:Each digit appears equally often in each place value position when all non‑repeating three‑digit numbers are formed.Explanation:Total numbers formed from digits $1$, $2$, $3$ without repetition is $3! = 6$.In these $6$ numbers, each digit occurs in the hundreds, tens, and units place exactly $2$ times (because $6 \div 3 = 2$).Sum contributed by the hundreds place: $(1+2+3) \times 2 \times 100 = 6 \times 2 \times 100 = 1200$.Sum contributed by the tens place: $(1+2+3) \times 2 \times 10 = 6 \times 2 \times 10 = 120$.Sum contributed by the units place: $(1+2+3) \times 2 \times 1 = 6 \times 2 \times 1 = 12$.Total sum = $1200 + 120 + 12 = 1332$.Alternatively, using the formula: sum = (sum of digits) $\times (n-1)! \times (10^0+10^1+\cdots+10^{n-1})$ gives $6 \times 2! \times 111 = 12 \times 111 = 1332$.Answer:$1332$