(i) Find the sum of 4-digited number that can be formed using the digits 1,2,3,4,5,6 without repetition.
(ii) Find the sum of 4-digited numbers that can be formed using the digits 1,3,5,7,9 without repetition.

No of 4 digigt number that can be formed using 1,2,4,5,6 is ${}^{5}P_4=120.$
In units place each digit occurs in ${}^{4}P_3$ time sum units places =${}^{4}P_3\left(1+2+4+5+6\right)$
The value of sum in units place $={}^{4}p_3(1+2+4+5+6)1$
The value of sum in units 10"s place $={}^{4}p_3(1+2+4+5+6)10$
The value of sum in units 100"s place $={}^{4}p_3(1+2+4+5+6)100$
The value of sum in units 1000"s place $={}^{4}p_3(1+2+4+5+6)1000$
sum of 4 digits number $={}^{4}p_3(1+2+4+5+6)\left(1+10+100+1000\right)$
                                          $\begin{array}{c}=4P_3\left(18\right)\left(1111\right)\\ =479952.\end{array}$
(ii) No of 4 digigt number that can be formed using 1,3,5,7,9without repetition is ${}^{5}P_4=120.$
In units place each digit occurs in ${}^{4}P_3$ time 
The value of sum in units place $={}^{4}p_3(1+3+5+7+9)1$
The value of sum in units 10"s place $={}^{4}p_3(1+3+5+7+9)10$
The value of sum in units 100"s place $={}^{4}p_3(1+3+5+7+9)100$
The value of sum in units 1000"s place $={}^{4}p_3(1+3+5+7+9)1000$
sum of 4 digits number $={}^{4}p_3(1+3+5+7+9)\left(1+10+100+1000\right)$
                                          $\begin{array}{c}=24\left(25\right)\left(1111\right)\\ =666600.\end{array}$