How many $3$-digit even numbers can be made using the digits $1,2,3,4,6,7$ if no digit is repeated?

A three digit even number is to be formed from given $6$ digits $1,2,3,4,6,7$.
$$\begin{gathered} \square \square \square \\ HTO \end{gathered}$$
Since, for the number is to be even , so ones place can be filled by $2,4$ or $6$. So, there are $3$ ways to fill ones place.
Since, repetition is not allowed , so tens place can be filled by remaining $5$ digits. So, tens place can be filled in $5$ ways.
Similarly, hundred's place can be filled by remaining $4$ digits. So, hundred's place can be filled in $4$ ways.
So,required number of ways in which three digit even numbers can be formed from the given digits is $4\times 5\times 3=60$

Alternative Method:
3-digit even  numbers are to be formed using the given six digits, $,2,3,4,6$ and $7$, without repeating the digits.
Then, units digits can be filled in $3$ ways by any of the digits, $2,4$ or $6$.
Since the digits cannot be repeated in the $3$-digit numbers and units place is already occupied with a digit (which is even), the hundreds and tens place is to be filled by the remaining $5$ digits.
 Therefore, the number of ways in which hundreds and tens place can be filled with the remaining $5$ digits Is the permutation of $5$ different digits taken $2$ at a time.
${}^5{P}_2=\frac{5!}{(5-2)!}=\frac{5!}{3!}$
Number of ways of filling hundreds and tens place
$=\frac{5\times 4\times 3!}{3!}=20$
Thus, by multiplication principle, the required number of $3$-digit numbers is $3\times 20=60$