Question

How many $3$ digit numbers can be formed with the digits $1,2,3,4,5$ where digits may not be repeated?
 $$\begin{gathered} A.60 \\ B.125 \\ C.216 \\ D.noneofthese \end{gathered}$$

Answer 1

Hint: To find the total number of ways in which three digit numbers can be formed with repletion. We first see that all given digits are different so to find total ways we first find the number of ways in which the unit digit can fill, then number of ways in which ten’s place fill and then the same for a hundred places and then multiplying all three results to get the total number of ways.

Complete step-by-step answer:
Here, we have to find a total number of $3$ digit numbers that can be formed by digits $1,2,3,4,5$ where repletion of digits is not allowed.
Since, it requires you to form a $3$ digit number but there are $5$ different digits available.
To find the total number of ways or numbers that can be formed. We discuss in how many ways each position of $3$ digit number can be filled.
Here, as there is no restriction on any given digit or any digit can come at any position.
So, at the unit place either of the given digit can come.
Hence, there are $5$ ways in which the unit digit of number can take its value.
Now, for tens place of number. We see that out given five digits one digit is already used at ones place.
Since, no repetition is allowed so for tens place only $4$ digits left from given $5$ digits.
And as there is no restriction on digits. So, any out of remaining $4$ can come at tens position of number.
Hence, there are $4$ ways in which ten’s place of number can be filled.
Now, for a hundred places of numbers we see that out of five digits two digits are already used. So, we left with only three digits and as there is no restriction on digits.
Hence, to fill a hundred places of a number there are $3$ different ways are there.
Therefore, from above we see that total number of ways in which a three digit number can be formed from given digits $1,2,3,4,5$ are given as:
 $$\begin{gathered} \Rightarrow 5\times 4\times 3 \\ =60ways \end{gathered}$$
Hence, the total number of $3$ digits numbers that can be formed from digits $1,2,3,4,5$ when repetition is not allowed are $60.$
So, the correct answer is “Option A”.

Note: We can also find the solution of a given problem by using the concept of combination instead of permutation. Here in combination we first make a selection of three digits out of five digits and then we make arrangements of three digits at three places to get the required solution of the problem. This can be done as:
 $$\begin{gathered} {}^5{C}_3{\times }^3{P}_3 \\ ={\displaystyle \frac{5!}{2!3!}}\times {\displaystyle \frac{3!}{0!}} \\ ={\displaystyle \frac{5\times 4\times 3!}{3!\times 2\times 1}}\times 3\times 2\times 1 \\ =10\times 6 \\ =60 \end{gathered}$$
Hence, from above we see that the result in this case is also the same.