Summing up!

We have to form numbers using 1,2,3,4 and 5.

If we fix any digit then we get $4!$ permutations for it. Say, we fix 1 on ten thousand's place and start moving right fixing other digits.

Then we get 4 choices for thousand's , 3 choices for hundred's , 2 choices for ten's and 1 choice for one's place. Thus total permutations possible for 1 at ten thousand's place is $4*3*2*1 = 24$ .

Similarly 1 is repeated 24 times on thousand's place , 24 times on hundred's place, 24 times on ten's place, and 24 times at one's place. No more permutations for the digit 1 are possible.Similarly each digit has same pattern.

Sum of face values at ten thousand's place is $24*(1+2+3+4+5) = 360$ For thousand's place also the sum of face value will be $360$ and so on .

Now the sum of place values shall be $\rightarrow$ $360*{10}^4 + 360*{10}^3 + 360*{10}^2 + 360*{10}^1 + 360 =$ $\boxed {3999960}$