How many even three-digit numbers have distinct digits and have no digit $5$? How many even three-digit numbers have distinct digits and have no digit $5$?
The answer my teacher gave was $252$, but I don't see how she got that. I thought it would be $6\times 8 \times 5=240$ because the $3\text{rd}$ digit must be even $(0,2,4,6,8)$, the $2\text{nd}$ digit can't be $5$ or the $3\text{rd}$ digit $(10-2=8$ options) and the first digit can't be $0$, $5$, the $2\text{nd}$ digit, or the $3\text{rd}$ digit ($10-4=6$ options). Any way I look at it, the last digit has to be even and there are $5$ options for even digits, so the final answer must end in a $5$ or a $0$, not a $2$. Please help!

No. ending with non-zero digit = $4 .7.7=196$.
No. ending with zero digit = $1 .8.7=56$
P.S. The product above is (third digit options) x (first digit) x (second digit)
