Find the sum of all three digit number having digits without repetition..

Hi,

First, let us consider a number having 1 as one of the digits. 1 will appear in the 100s place 9*8 = 72 times.

It will appear in the 10s place 8*8 = 64 times and in 1s place for 64 times.

This holds true for all numbers as well.

Hence, sum of all 3 digit numbers = 72*100(sum of 1 to 9) + 64*10(sum of 1 to 9) + 64* 1(sum of 1 to 9)

= 45*(7200+640+64) = 355680