The sum of five digit numbers which can be formed with the digits 3, 4, 5, 6, 7 using each digit only once in each arrangement, is

The number of numbers of five digits, having 3 in unit’s place ${=}^4{\mathrm{P}}_4$ (number of arrangements of the remaining four digits)

= 4! = 24.

Similar is the case for other digits, i.e., each digit occurs 24 times in the unit’s place.

Hence, the sum of the digits in unit’s place

$\begin{array}{l}=24\times 3+24\times 4+24\times 5+24\times 6+24\times 7\\ =24(3+4+5+6+7)=24\times 25=600\end{array}$

Similarly, the sum of the digits in each of the other places = 600.

∴ The required sum

$$\begin{gathered} =600\times 10000+600\times 1000+600\times 100+600\times 10+600\times 1 \\ =600(10000+1000+100+10+1) \\ =600\times 11111=6666600. \end{gathered}$$