Combinatorics: throwing a dice three times to get an even number. Suppose you throw a six face dice three times, how many times will be the sum of the faces even?
I approached it this way:
You either get all three times even face, or twice odd and once even.
As there are only 3 faces that are even, you have $3^3$ possibilities.
Then, for the second situation, you have $3$ choices for then face, twice $3$ choices for the odd one. Thus giving again $3^3$ possibilities.
Overall, there will be $3^3 + 3^3$ possibilities, yet my textbook shows $3^3 + 3^4$ possibilities. What's wrong with my reasoning?

P(all even)$=\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)=\dfrac{1}{8}$
P(one even and two odd)$=\dbinom{3}{2}\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)=\dfrac{3}{8}$
Required probability $=\dfrac{1}{8}+\dfrac{3}{8}=\dfrac{1}{2}$
