Find the sum of all $4-digit$ numbers that can be formed using the digits $0,2,4,7,8$ without repetition.

Solution:

The number of $4-digit$ numbers formed by using $0,2,4,7,8$ without repetition $={}^5{P}_4-{}^4{P}_3=120-24=96.$

Out of these $96$ numbers,

${}^4{P}_3-{}^3{P}_2$ numbers contain $2$ in units place.

${}^4{P}_3-{}^3{P}_2$ numbers contain $2$ in tens place.

${}^4{P}_3-{}^3{P}_2$ numbers contain $2$ in hundreds place.

${}^4{P}_3$ numbers contain $2$ in units place.

$\therefore$ The values obtained by adding $2$ in all numbers.

$=({}^4{P}_3-{}^3{P}_2)2+({}^4{P}_3-{}^3{P}_2)20+({}^4{P}_3-{}^3{P}_2)200+{}^4{P}_2\times 2000$

$={}^4{P}_3(2+20+200+2000)-{}^3{P}_2(2+20+200)$

$=24\times 2222-6\times 222$

$=24\times 2\times 1111-6\times 2\times 111$

Similarly, the value obtained by adding $4$ is $24\times 4\times 1111-6\times 4\times 111.$

The value obtained by adding $7$ is $24\times 7\times 1111-6\times 7\times 111.$

The value obtained by adding $8$ is $24\times 8\times 1111-6\times 8\times 111.$

$\therefore$ The sum of all numbers

$=(24\times 2\times 1111-6\times 2\times 111)+$ $(24\times 4\times 1111-6\times 4\times 111)+$ $(24\times 7\times 1111-6\times 7\times 111)+$ $(24\times 8\times 1111-6\times 8\times 111)$

$=24\times 1111\times (2+4+7+8)-6\times 111\times (2+4+7+8)$

$=26664\times 21-666\times 21$

$=559944-13986$

$=545958$