Question

How many 3-digit numbers can be formed using the digits 1,2,3,4,5,6 without repeating any digit?

Answer 1

Hint: We have to find the total number of 3 digit numbers using the digits 1,2,3,4,5,6 without repeating any digit. For this, we will count how many digits can be put in the hundredth’s place and then how many can be put in the tens’ place and then how many can be put in the ones’ place. After that, we will multiply all these three numbers and we will get our answer.

Complete step-by-step solution:
Now, starting from the hundredth’s place in the number,
We have been given the digits 1,2,3,4,5,6 which means the number of available digits we have is 6.
Since no digit out of the given six digits has been used yet, the hundredth’s place can be filled in a total of ‘6’ ways.
Now, for the tens’ place in the number,
Now there’s one digit used out of the six on the hundredth’s place. Since the repetition of digits is not allowed that digit cannot be at the tens’ place. For example, if the number starts from ‘4’ neither the tens’ nor the ones’ place can have ‘4’ in it.
Therefore, there are $'6-1=5'$ ways in which the tens’ place can be filled.
Now, for the ones’ place in the number,
There are now 2 digits used out of the six given digits by the hundredths and tens’ place and since there cannot be any repetition we cannot use either of those two digits at the ones’ place.
Therefore, there are $'6-2=4'$ ways in which the ones’ place can be filled.
Now, we have established that the hundredth’s place can be filled in ‘6’ ways, the tens’ in ‘5’ ways, and the ones’ in ‘4’ ways.
Thus, as mentioned in the hint the total number of 3 digit numbers formed by using the digits 1,2,3,4,5,6 without repeating any of them is given by:
$\begin{aligned} & \Rightarrow 6\times 5\times 4 \\\ & \Rightarrow 120 \\\ \end{aligned}$
Thus, there are a total of 120 numbers that can be formed by using the digits 1,2,3,4,5,6 without repeating any digit.

Note: This question can also be done by the following method:
We will here use the $^{n}{{P}_{r}}$ operator to find out the required number of 3 digit numbers.
To form 3 digit numbers out of the digits 1,2,3,4,5,6 we need to select 3 different digits from these 6 digits and arrange them in all possible ways. The total number of ways in which this selection and arrangement can be done will be equal to the required number of 3 digit numbers.
When we have to find out the number of ways of both selection and arrangement simultaneously, we use the $^{n}{{P}_{r}}$ operator where ‘n’ is the total number of objects out of which selection and arrangement have to be done and ‘r’ are the total number of objects which have to selected and arranged out of the ‘n’ given objects.
Here,
$\begin{aligned} & n=6 \\\ & r=3 \\\ \end{aligned}$
And we know that $^{n}{{P}_{r}}$is given as:
$^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$
Therefore, the required number of 3 digit numbers is given as:
$\begin{aligned} & {{\Rightarrow }^{6}}{{P}_{3}} \\\ & \Rightarrow \dfrac{6!}{\left( 6-3 \right)!} \\\ & \Rightarrow \dfrac{6!}{3!} \\\ & \Rightarrow 120 \\\ \end{aligned}$
Hence, the required number of 3 digit numbers is 120.