How many six-digit numbers can be formed from the digits $1$, $2$, $3$, $4$, $5$, $6$ and $7$ so that the digits do not repeat and the terminal digits should be even.
Given: $\;$ Three even digits -- $2$, $4$ and $6$
There are two terminal digits in a six digit number.
$2$ terminal digits from given $3$ even digits can be selected in
$P^3_2 = \dfrac{3!}{\left(3-2\right)!} = 3! = 6$ ways
Since the digits are not repeated, the remaining $4$ digits of the number can be selected from the remaining given $5$ digits in
$P^5_4 = 5! = 120$ ways
$\therefore \;$ Total number of six-digit numbers that can be formed from the given digits so that the digits are not repeated and the terminal digits are even are
$6 \times 120 = 720$ numbers