How many $3$ digit even numbers can be formed by using digits $1,2,3,4,5,6,7$, if no digits are repeated? How many $3$ digit even numbers can be formed by using digits $1,2,3,4,5,6,7$, if no digits are repeated?
ATTEMPT
There are three places to be filled in _ _ _
I wrote it like this 
_ _ $2$
_ _ $4$
_ _ $6$
Now each of the two blanks can be filled in $P(6,2)$ ways. So adding results of three cases i have $3$. $P(6,2)$ which gives me $90$ ways. But the textbook states $60$ ways. Can someone suggest how ?
Thanks. 

For the first digit you only have $3$ possibilities, $2,4$ or $6$. For the second digit, having chosen the first digit you only have $6$ possibilities left. For the third digit, having chosen the first two digit you have $5$ possibilities left. In total you have $3 \times 6 \times 5$ possibilities.
