How many 3 digit numbers can be formed from the digits 0, 1, 2, 3 and 4 with repetitions?
Hint: In this question the rightmost digit of the 3 digits number can be chosen from 5 digits that is 0, 1, 2, 3 and 4, now the left to this digit or the tenth place can be again chosen from the same 5 digits, now the hundredth place digit can only be chosen from 4 digits as 0 can’t be the first digit of a number.
Complete step-by-step answer: Given digits (0, 1, 2, 3 and 4) So the total number of digits is 5 Now we have to make a 3 digit number with repetitions. $ \bullet \bullet \bullet $ So as we see the unit place is filled in 5 ways. Now the ten’s place is also filled by 5 ways as repetition is allowed. Now the hundreds place is filled by 4 ways because digit 0 cannot be placed at hundred’s place otherwise the three digit numbers are converted into 2 digit numbers for example (012 = 12) and the repetition is allowed so the hundreds place is filled by 4 ways. So the total number of ways to make a three digit number from the given digits and repetition is allowed are: Total number of ways = $4 \times 5 \times 5$ = $100$ So the total number of ways to make a three digit number from the given digits and repetition is allowed are 100. So this is the required answer.
Note: In this question it’s provided with repetition, this means that the digits once used are available to be used again however if this would not have been the case then the unit’s place could have been chosen from 5 digits that is 0, 1, 2, 3 and 4, then the tenth digits should have been chosen from 4 digits as 1 digits is already being used for the unit’s place and then the hundredth place could have been chosen from 2 digits as 2 digits are already been chosen previously and we can’t chose 0 as the first digits of a number thus the answer in that case would have been $2 \times 4 \times 5$.