Find the sum of all of the three digit positive integers with three distinct digits.

Let us consider a number having 1 as one of its digits. Then, clearly the 100s place will contain 9*8=72 ones. The 10s , and the unit place will each consist of 8*8=64 1s

Threfore, the required sum:

= 7200( sum of 1 to 9 ) + 640(sum of 1 to 9) +64( sum of 1 to 9)

= 45(7200+640+64)

=355,680

*Edited on ***March 28, 2024, 9:03 am**