Question

How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is not allowed?

Answer 1

Hint:Consider three spaces, find the number of ways in which each of these $3$ spaces can be filled without the digits repeating. Then multiply the number of ways for each space to get a total number of $3$ digit numbers possible.

Complete step-by-step answer:
Given the digits 1, 2, 3, 4 and 5 with these we need to form a 3 digit number;
\[\begin{matrix}
   \overline{\downarrow } & \overline{\downarrow } & \overline{\downarrow } \\
   {{1}^{st}} & {{2}^{nd}} & {{3}^{rd}} \\
   digit & digit & digit \\
\end{matrix}\]
We can fill the 1st position in 5 ways as we are given a total of 5 digits.
We can fill the 2nd position in 4 ways, because repetition is not allowed. The 2nd place can be filled in 4 ways only.
We can fill the 3rd position in 3 ways, because repetition is not allowed, so 3rd place can be filled in 3 ways;
\[\begin{matrix}
   5ways & 4ways & 3ways \\
   \underline{\downarrow } & \underline{\downarrow } & \underline{\downarrow } \\
   {{1}^{st}} & {{2}^{nd}} & {{3}^{rd}} \\
\end{matrix}\]
$\therefore $ Number of 3 digit number without repetition =5 X 4 X 3 = 20 X 3 = 60
$\therefore $ Number of three digit number possible = 60

Note: If in the case that repetition of digits is allowed then;
\[\begin{matrix}
   5ways & 5ways & 5ways \\
   \underline{\downarrow } & \underline{\downarrow } & \underline{\downarrow } \\
   {{1}^{st}} & {{2}^{nd}} & {{3}^{rd}} \\
\end{matrix}\]
The place can be filled with repetition;
$\therefore $Number of 3 digit number possible = 5 X 5 X 5 = 125.