Question

number of 3 digit numbers having at least one of their digit as 5 are​

Answer 1

Answer:

Case1

5 comes at hundreds place

So now we have 9(0,1,2,3,4,6,7,8,9)options to fill the tens place and 9 places to fill ones place

So 1×9×9=81 ways

Case2

5 comes at tens place

Now hundreds place can be filled with 8 options (1,2,3,4,6,7,8,9 not zero as then it will become a 2 digit number) and 9 ways to fill ones place

So 8×1×9=72

Case 3

5 comes at ones place

So again 8 ways to fill hundreds place and 9 ways to fill tens place

8×9×1=72

So total ways =81+72+72= 225 ways

Hope it helps

Step-by-step explanation:

Case 1 If the unit digit is 5 then

We have 8*9*1=72numbers

Case 2 If tens digit is 5 then

We have 8*1*9 =72 numbers

Case 3 When hundreds digit is 5 then

We have 1*9*9=81 numbers

Total three digit numbers having one or more 5 are 72+72+81=225 numbers.