Hint- Here, we will proceed by using the concept of permutations. We will find out exactly how many digits out of the digits 0,1,2,3,4,5,6,7,8,9 can be occupied at the one’s place and ten’s place in order to obtain any 2-digit number.
Complete step-by-step solution - As we know any 2-digit number consists of two digits placed at its one’s place and ten’s place and these digits can be 0,1,2,3,4,5,6,7,8,9 (total 10 digits). For any number to be a 2-digit number, 0 should not occur at the ten’s place because if 0 occurs at the ten’s place this number will become a 1-digit number. So, any one of the ten digits (0,1,2,3,4,5,6,7,8,9) can be placed at the one’s place and any one of the nine digits (1,2,3,4,5,6,7,8,9) can be placed at the ten’s place in order to get a 2-digit number. Number of possible digits which can be placed at one’s place = 10 Number of possible digits which can be placed at ten’s place = 9 Since, Total number of 2-digit numbers = (Number of possible digits which can be placed at one’s place)$ \times $( Number of possible digits which can be placed at ten’s place) $ \Rightarrow $Total number of 2-digit numbers = $10 \times 9 = 90$ Therefore, the total number of 2-digit numbers which can be obtained are 90.
Note- In this particular problem, we have used the basic method of finding out the total number of n-digit numbers where n can be 1,2,3, etc. The total number of 2-digits numbers which can be obtained are easily found by simply counting as numbers 0 to 9 are 1-digit numbers and the numbers 11 to 99 are 2-digit numbers which are 90.