How many even numbers could be formed

Question

How many even numbers of three different digits less than $500$ can be formed from the integers $1, 2, 3, 4, 5, 7$ and $8$?

Answer 1

For numbers with the hundreds digit $=4$, there are only two other even digits remaining ($2,8$). So the units place has two choices available. After choosing the units digit, there are $5$ remaining digits to choose from for the tens place. So there are $1\cdot 5\cdot 2=10$ even three digit numbers beginning with $4$. A similar analysis applies to numbers beginning with the digit $2$, giving another $10$ possibilities. As to numbers beginning with any of the odd digits, we see that there are $2$ possibilities for the hundreds place digit, $3$ possibilities for the units digit, thereafter $5$ possibilities for the tens digit yielding $2\cdot 5\cdot 3=30$. So the total number of possibilities is $10+10+30=50$.

Answer 2

100th digit must be $1,2,3,4$, every digit is different. So $$4*6*5=120$$