Analyse it carefully

Basically what we need to do is calculate the sum of all the numbers that can be formed from the set { $1,2,3,4,5$ } , taken all at a time .

Total number of numbers that can be formed is $5! = 120$ .

So to calculate the sum of all the possible permutations , we use the formula : $(n-1)! \cdot (\text{Sum of all the digits used}) \cdot ( 10^{0} + 10^{1} + \dots + 10^{n-1}) \\ 4! \cdot 15 \cdot 1111 \\=399960$

So $x$ evaluates to be 9 .

$\bf{How the Formula Works}$ :

( $n$ is the total number of digits to be used )

Each of the $n$ digits used , occur a total $(n-1)!$ times in all the possible numbers , in each of the $10^{i th}$ place

Now sum of digits in each of the $10^{i th}$ place is $(\text{Sum of the digits}) \cdot (n-1)!$ .

Now adding the sum from all the $10^{i}$ places , we get the above used formula $(n-1)! \cdot (\text{Sum of all the digits used}) \cdot ( 10^{0} + 10^{1} + \dots + 10^{n-1})$ .

there are 120 different numbers which can be formed by using the numbers 1 2 3 4 5(without repeating). so there would be 120 times 1, 120 times 2, 120 times 3, 120 times 4 and 120 times 5.so the digit sum of the sum of all the permutations of 12345 will be equal to 120 * (1 + 2 + 3 + 4 + 5). so the sum is 1800. we once more find the sum which is equal to 9.

....

Please inform me if there is any error in my reasoning. It's just a logic, I have no proof for it. Submit it if you have.