Find the sum of all the number formed by 2,4,6, and 8 without repetition.Number may be of any digit like 2, 24, 684, 4862.

There are:

• four $1$-digit numbers, whose average is $5$: sum = $20$
• twelve $2$-digit numbers, whose average is $55$: sum = $660$
• twenty-four $3$-digit numbers, whose average is $555$: sum = $13320$
• twenty-four $4$-digit numbers, whose average is $5555$: sum = $133320$

To see why these averages are correct, you can pair off each number uniquely with its 'complement' obtained by exchanging $2$ with $8$ and $4$ with $6$. The average of each of these pairs is $5\dots5$.

To sum all the numbers with four digits, note that for any given digit in any given position, it appears in the sum six times. That is, for example, there are six ways to complete a number if we know that its third digit is $4$. Then, the sum is $$6(2222+4444+6666+8888)=6(2+4+6+8)1111$$