Question

How many 3 digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that
(I) Repetition of the digits is allowed?
(II) Repetition of the digits is not allowed?

Answer 1

Hint: In this question, we are given 5 digits and we have to form a 3 digit number for the given two conditions. For this, we will use the concept of permutation. We will find the number of digits that can be placed at the three places for each part. Then multiply ways for each part to get the answer for each part.

Complete step by step answer:
Here we are given 5 digits from which we need to form a 3 digit number. We know that the place for numbers is ones, tens, and hundreds. Now let us find ways for each part.
(I) If repetition of the digits are allowed.
Here we have digits 1, 2, 3, 4, and 5 and we can repeat them to form a 3 digit number.
For one’s place, we can have any of the 5 digits. So the number of ways for placing a digit in the one’s place is 5.
For tens place, we can have any of the 5 digits as repetition is allowed. So the number of ways for placing a digit at tens place is 5.
For the hundreds place, we can have any of the 5 digits as repetition is allowed. So the number of ways for placing a digit at the hundreds place is 5.
\[\dfrac{5\text{ ways}}{\text{Hundreds}}\dfrac{5\text{ ways}}{\text{Tens}}\dfrac{5\text{ ways}}{\text{Ones}}\].
Since all events are occurring simultaneously (the number is formed by placing all the digits), so we will multiply the ways to get total ways.
Hence total ways becomes $ 5\times 5\times 5=125\text{ ways} $ .
(II) If repetition is not allowed.
For one’s place, we have all 5 choices. So the number of ways for placing a digit at the one’s place is 5.
For tens place, we have 4 choices left because repetition is not allowed and one digit is already placed at the one’s place. So the number of ways for placing a digit at tens place is 4.
For the hundreds place, we are left with 3 choices because 2 digits are already placed at ones and hundreds place. So the number of ways for placing a digit at the hundreds place is 3.
\[\dfrac{3\text{ ways}}{\text{Hundreds}}\dfrac{4\text{ ways}}{\text{Tens}}\dfrac{5\text{ ways}}{\text{Ones}}\].
Hence total number of ways $ 5\times 4\times 3=60\text{ ways} $ .

Note:
Students should consider every possibility while calculating the number of ways. In case, one of the digits was 0, then we could not place it as the hundreds place. Take care of the condition for repetition if it is allowed or not.