We wish to find how many even numbers less than $600$ can be formed from the digits $3, 3, 4, 8, 9$ if each digit is used at most once.
Since the number is even, the units digit of each number must be $4$ or $8$.
One-digit numbers: The only possibilities are $4$ or $8$, giving us two possibilities in this case, as you found.
Two-digit numbers: If the units digit is $4$, then the tens digit can be $3$, $8$, or $9$. If the units digit is $8$, then the tens digit can be $3$, $4$, or $9$. Hence, there are six possibilities, as you found.
Three-digit even numbers: If the units digit of the even number less than $600$ is $4$, the hundreds digit must be $3$. This leaves us with three choices for the tens digit, namely $3$, $8$, or $9$. Hence, we can form three three-digit even numbers less than $600$ with units digit $4$ by using the digits $3, 3, 4, 8, 9$ at most once. They are $334$, $384$, $394$.
If the units digit of the three digit even number less than $600$ is $8$, we have two possibilities for the hundreds digit, namely $3$ or $4$. If the hundreds digit is $3$, we have three possibilities for the tens digit, namely $3$, $4$, or $9$. If the hundreds digit is $4$, we have two possibilities for the tens digit, namely $3$ or $9$. Thus, we can form five three-digit even numbers less than $600$ with units digit $8$. They are $338$, $348$, $398$, $438$, $498$.
Hence, there are a total of eight three-digit even numbers less than $600$ that can be formed with the digits $3, 3, 4, 8, 9$ if each digit is used at most once.
In all, we can form $2 + 6 + 8 = 16$ even numbers less than $600$ using the digits $3, 3, 4, 8, 9$ at most once.
Best Answer
The hints in the comment section provides valuable insights to the problem.
I will write a complete answer just to be sure you understand the hints.
There are $3$ cases whereby the sum of the $4$ digits is even,
(1) All $4$ digits are odd.
(2) All $4$ digits are even.
(3) $2$ digits are odd and $2$ digits are even.
First, note that there are $4$ odd digits and $3$ even digits in the set $\{1,2,3,4,5,6,7\}$.
Let's start with Case (1).
Since each digit can only be chosen once, there is $4$ ways of choosing the first odd digit from $\{1,3,5,7\}$,
followed by $3$ ways of choosing the second odd digit from the remaining $3$ digits,
followed by $2$ ways of choosing the third odd digit from the remaining $2$ digits,
followed by $1$ way of choosing the forth odd digit from the remaining $1$ digit.
So, there are $4\times 3 \times 2\times 1 = 4!$ possibilities for Case (1).
Now on to Case (2),
there are $3$ even digits in $\{1,2,3,4,5,6,7\}$, namely $2,4,6$.
Remember that the $4$ digit number must contain different digits, we cannot use either $2$, $4$ or $6$ more than once to construct the $4$ digit number.
And furthermore, there are only $3$ even digits available, so it is impossible to construct any $4$ digit number with no repetition of a digit.
Hence, there is $0$ possibilities for Case (2).
Finally, let's move on to Case (3), which is slightly more difficult.
We first do an unordered selection of $2$ odd digits from $\{1,3,5,7\}$.
which gives $\binom{4}{2}$ possibilities,
followed by an unordered selection of $2$ even digits from $\{2,4,6\}$.
which gives $\binom{3}{2}$ possibilities.
An example would be choosing the digits $1, 5, 2, 4$.
We can quickly see that a valid $4$ digit number would be $1524$.
Note that other permutations of the $4$ digits also form a valid $4$ digit number,
example, $5124$ and $1542$.
and, there are $4!$ permutations in total.
So, by the product rule, there are $\binom{4}{2} \times \binom{3}{2} \times 4!$ possibilities.