Question

Find the number of 4-digit even numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of those will be even? \[\]

Answer 1

Hint: We find number of ways we can fill the one’s, tenth, hundredth, thousandth place of a 4-digits number from the given digits 1, 2, 3, 4, 5 and use rule of product to get the total number of four digit numbers. We fill the one’s place with even digits and fill the rest of the places like we filled for total 4-digit numbers to find number of total even 4-digits numbers.\[\]

Complete step by step answer:
We know that a 4-digit number has four places: one’s place, ten’s place, hundred’s place and thousandth place.
\[\begin{matrix}
   \_ & \_ & \_ & \_ \\
   Th & H & T & O \\
\end{matrix}\]
So we have to fill four places without repetition. We have 5 digits 1, 2, 3, 4 and 5. We fill the one’ place with 5 ways with one of the digits choosing from 1, 2, 3, 4, 5. We then fill ten’s place with 4 ways with remaining 4 digits since one digit has been already used for one’s place. We fill the hundredth place in 3 ways with remaining 3 digits and the thousandth place in 2 ways with remaining 2 digits. So by rule of product the number of ways ; we can make 4-digit numbers from digits 1, 2, 3, 4, 5 is;
\[2\times 3\times 4\times 5=120\]
If the number has to be even, the one’ place has to be filled with even digits. So we can only fill one’s place with digits 2 or 4 that is in 2wyas. We then fill ten’s place with 4 ways with remaining 4 digits since one even digit has been already used for one’s place. We fill the hundredth place in 3 ways with remaining 3 digits and the thousandth place in 2 ways with remaining 2 digits just like we solved before. So by rule of product the number of ways; we can make 4-digit even numbers from digits 1, 2, 3, 4, 5 is;
\[2\times 3\times 4\times 2=48\]

Note: We note that the rule of product states that if there are $m$ ways to do something and $n$ ways to other things then there are $m\times n$ ways to do both things. We can alternatively solve using the permutation that is the number of arrangements of $n$ distinct objects in $r$ fixed slots or position is called “$r$ permutation of $n$”, denoted as ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$.