Question

Find the sum of all $4$ digit numbers that can be formed using the digits $1,3,5,7$ and $9$ (without repetition).

Answer 1

The number of arrangement of $4$ different digits taken $4$ at a time is given by ${}^4{P}_4=4!=24$.

All the four digits will occur equal number of times at each of the positions, namely ones, tens, , hundreds, thousands.

Thus, each digit will occur $24/4=6$ times in each of the positions. The sum of digits in one's position will be $6\times (1+3+5+7)=96$.

Similar is the case in ten's hundred's and thousand's places.

Therefore, the sum will be $96+96\times 10+96\times 100+96\times 1000=106656$.