2023 CSAT — Q47

“$4$-digit, less than $2000$, digits $1,2,3,4$ unrepeated” forces the thousands digit to be $1$ (a $2,3,$ or $4$ there would exceed $2000$). The remaining three places are a permutation of $\{2,3,4\}$, giving $3! = 6$ numbers.

Thousands contribution: the digit $1$ appears in all $6$ numbers: $1 \times 1000 \times 6 = 6000$.

Other three places: each of $2,3,4$ appears in each of the hundreds/tens/units positions equally — $6$ numbers $\div\ 3$ digits $= 2$ times per digit per position. Per position the digit-sum contribution is $(2+3+4)\times 2 = 9 \times 2 = 18$ digit-units. So:

• hundreds: $18 \times 100 = 1800$
• tens: $18 \times 10 = 180$
• units: $18 \times 1 = 18$

Total:

Answer: (a) 7998.

The constraint "$< 2000$" is the lever — it fixes the thousands digit at $1$, cutting $24$ permutations to $6$. Then use positional frequency: with the lead fixed, each remaining digit visits each lower position $6/3 = 2$ times. Summing by place value ($6000 + 1800 + 180 + 18$) is far faster and safer than adding six four-digit numbers by hand. The decoy $9238$ is the all-permutations sum that ignores the cap.