Sums of Permutations

The possible permutations are 120 ... and there are 5 numbers and 4 places...So every number will repeat 24 times in each place ( units place , tens place , etc ) ... Now we can sum each place like [ 24 ( 1+2+3+4+5) ] and add the rest to the other place until we have the complete number and it will be 399960 Sorry For bad language :)

Smallest correct solution -

As the mean digit in each column is 3, each number is 3333, on average. Hence the sum is 120 * 3333 = 399,960.

$The ~sum ~of ~permutations ~of ~all$ $4$ $digit~ no.s~ using~ digits$ $1,2,3,4 = [1111 \times {6 \times (1+2+3+4)}] = 66660$

$similarly~ sum ~of ~permutations~ using$ $1,2,3,5 = [1111 \times {6 \times (1+2+3+5)}] = 73326$

$\dots \dots \dots$ $using$ $1,2,4,5 = [1111 \times {6 \times ( 1+2+4+5)}] = 79992$

$\dots \dots \dots$ $using$ $1,3,4,5 = [1111 \times {6 \times (1+3+4+5)}] = 86658$

$\dots \dots \dots$ $using$ $2,3,4,5 = [1111 \times {6 \times (2+3+4+5)}] = 93324$

$therefore ~total~ sum ~of~ all~ permutations = 399960$

Let N=abcd=1000a+100b+10c+d

(I am using £ as summation sign)

£N=1000£a + 100£b + 10£c + £d

£d = sum of all last digits in all permutations= 4P3(1+2+3+4+5)= 360

Similarly £a=£b=£c=360

£N=360(1000+100+10+1)=399960