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. 
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.

My Approach:
single digit no formed = 2,4,6,8
sum= 2+4+6+8= 20
two digit= 24,26,28,42,46,48,62,64,68,82,84,86
sum= 660
three digit no=246+264+426+462+624+642=2664
268+286+628+682+826+862=3552
248+284+428+482+824+842=3108
468+486+648+684+846+864=3996
sum of all 3 digit nos =13310
Similarly for all 4 digit numbers.
 A: 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 $5\dots5$.
A: 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
$$6(2222+4444+6666+8888)=6(2+4+6+8)1111$$
A: I guess this might help.
sum of number by taking $n$ digits all at a time $= (n-1!)(x_1+ x_2 + x_3 + \cdots + x_n)$
so if we want to find the sum of all the number formed by $2,4,6$ and $8$ without repetition, "without repetition".
[
  ( 2 + 4 + 6 + 8 )( 3! ) = 120.
]
Now multiply this with the place values = $120 ( 10^0 + 10^1 + 10^2 + 10^3 )$
if we want to find the sum of all the number formed by $2,4,6$ and $8$ without repetition, "with repetition", then
$( 2 + 4 + 6 + 8 )( 4*4*4 )$ (keeping $2$ at units place _ _ _ $2$), and the next step is same.
A: Simple formula :- (n-1)! * Sum of digits * 1111 n times.
That is, (4-1)! * (2+4+6+8)* 1111
= 6*20*1111
= 1,33,320
A: Consider an example by taking digits as $3,4,6,8$.
total combinations possible$=4\times 3\times 2=24$ numbers possible
no of digits$=4$
hence $\dfrac{24}{4}$ each digit comes $6$ times in ones, tens and hundred's place
at ones place=$6\times (3+4+6+8)=126\equiv6$.
at ten's place$=6\times (3+4+6+8)+12(\text{carry})=138\equiv 8$.
At hundred's place$=6\times (3+4+6+8)=139\equiv 9$
so, $3$ will come at thousands place and $1$ at ten thousands place,
Hence, the number is $139864$.
