JEE Main 2023 — Mathematics Algebra

Total $4$4 digit numbers using the digits $1,2,2,3$1,2,2,3 is

$$=\frac{4!}{2!}=12$$=2!4!​=12

Numbers are

___3 ___3 ___3 ___2 ___2 ___2 ___2 ___2 ___2 ___1 ___1 ___1

Required sum is

$$=\underset{\mathrm{unit}\mathrm{digit}'s\mathrm{sum}}{\underset{⏟}{[3\times 3+2\times 6+1\times 3]}}+\underset{\mathrm{ten}'s\mathrm{place}\mathrm{sum}}{\underset{⏟}{[(3\times 3+2\times 6+1\times 3)\times 10]}}+\underset{\mathrm{hundred}'s\mathrm{place}\mathrm{sum}}{\underset{⏟}{[(3\times 3+2\times 6+1\times 3)\times 100]}}+\underset{\mathrm{thousand}'s\mathrm{place}\mathrm{sum}}{\underset{⏟}{[(3\times 3+2\times 6+1\times 3)\times 1000]}}$$=unitdigit′ssum⏟[3×3+2×6+1×3]​​+ten′splacesum⏟[(3×3+2×6+1×3)×10]​​+hundred′splacesum⏟[(3×3+2×6+1×3)×100]​​+thousand′splacesum⏟[(3×3+2×6+1×3)×1000]​​

$$=[24]+[24\times 10]+[24\times 100]+[24\times 1000]$$=[24]+[24×10]+[24×100]+[24×1000]

$$=24+240+2400+24000=26664$$=24+240+2400+24000=26664