The sum of all the four-digit numbers that can be formed using all the digits 2, 1, 2, 3 is equal to \_\_\_\_\_\_\_\_\_\_.
Answer: 26664
Solutions · 1
Official solution
Digits: 2, 1, 2, 3 (with 2 repeated). Total distinct arrangements: 2!4!=12.
For a fixed digit in a given position (say unit place), count how many arrangements place each digit there:
- Digit 1 at unit place: remaining 2,2,3 arranged in 2!3!=3 ways
- Digit 2 at unit place: remaining 1,2,3 arranged in 3!=6 ways
- Digit 3 at unit place: remaining 1,2,2 arranged in 2!3!=3 ways
Sum of digits at each place: 3(1)+6(2)+3(3)=3+12+9=24.
Since this holds for each of the 4 place values:
\[\text{Sum} = 24\times(1000+100+10+1) = 24\times1111 = 26664\]