# 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.

• In each you have, altogether, $5\cdot 4\cdot 3$ possibilities, or $60$. Jun 1, 2015 at 15:42
• I think that your textbook is wrong. I also get $90$ numbers. Jun 1, 2015 at 15:44
• $60$ is correct if you only have numbers $(1,2,3,4,5,6)$ to chose from. Or numbers $(2,3,4,5,6,7)$, Jun 1, 2015 at 23:46

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.