sum of 4 digit nos. formed by taking 4,8,2,6

This is the case of permuting the digits without repeating any of them.

Consider a simple case of 1, 2, 3. The permutations of these are 123, 132, 231, 213, 312, 321.

There are a total of 3! = 6 permutations. The number of times that 1 appears in the units place = (3-1)! = 2!. Similarly, it appears 2! times in the tens and hundreds place respectively.

So, the summation for 1 is: 2!*100 + 2!*10 + 2!*1 = 111*2!

The summation for 2 is: 2!*200 + 2!*20 + 2!*2 = 2!*222 = 2!*111*2

The summation for 3 is: 2!*300 + 2!*30 + 2!*3 = 2!*333 = 2!*111*3

Total sum = 111*2!*(1+2+3) = 111*2!*6

In general, the formula for 'n' digits is (n-1)!*1111... (n times)*(sum of the digits)

Sum of the numbers = 1111*3!*(4+8+2+6) = 1111*120 = 133320

explanation?

Why 3! ?