The sum of 4 digited numbers formed by using the digits 0, 2, 4, 7, 8 without repetition is

Sum of 4 digited numbers is given by the formula (involving zero)

Sum of numbers $=\left[{}^{\left(n-1\right)}{P}_{\left(r-1\right)}\times \text{sum of digits}\times \left(1111.1\right)\text{'r' times}\right]$

$-\left[{}^{n-2}{P}_{r-2}\times \text{sum of digits}\times 111.1\left(r-1\right)\text{times}\right]$

Here $n=5,r=4$

$\therefore$  Sum of numbers $=\left[{}^4{P}_3\times 21\times 1111\right]-\left[{}^3{P}_2\times 21\times 111\right]$

$=21\times \left[{}^4{P}_3\times 1111{-}^3{P}_2\times 111\right]$

$=545958$