bmwhype2 wrote:

What is the SUM of all possible 3 digit numbers that can be constructed using the digits 3,4,5 and if each digit can be used only one in each number?

Can someone explain this in detail?

this is a good question.

There can be six numbers with 3 4 and 5 with no repetitions.

3!.

now in each place i.3 units, tens and hundreds each number would show up twice. So the total would be 2*(3+4+5) = 24.

So while adding the units place we get 24 so we keep 4 and 2 carries over.

while adding tens place we get 24 + 2 = 26 so we keep 6 and 2 carries over. While adding hundred's place we have 24 + 2 = 26.

so the sum is 2664 as the answer.