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.
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.
Finally, by sum rule, there are a total $4! + 0 +\binom{4}{2} \times \binom{3}{2} \times 4!$ possibilities for the $3$ mutually exclusive cases.
Best Answer
Note: I initially solved the problem using tedious case work (Method 2). I then figured out a more efficient method (Method 1) in which I considered whether or not the digit $3$ is used twice. I decided to leave Method 2 as a check on Method 1.
Method 1: We consider whether or not the digit $3$ is used twice.
The requirement that the number must be odd means the units digit is either $3$, $5$, or $7$.
Case 1: The digit $3$ is used at most once. Hence, we are selecting from the set $\{2, 3, 5, 6, 7\}$ without repetition.
One-digit numbers: There are three possible single-digit odd numbers, namely $3$, $5$, and $7$.
Two-digit numbers: There are three choices for the units digit. For each such choice, there are $4$ choices for the tens digit (any number other than the units digit). Therefore, there are $4 \cdot 3 = 12$ such numbers.
Three-digit numbers: The requirement that the number be less than $600$ means the hundreds digit must be less than $6$.
If the units digit is $7$, there are three choices for the hundreds digit ($2$, $3$, or $5$) and three choices for the tens digit (since we must exclude the units digit and the hundreds digit). Thus, there are $3 \cdot 3 = 9$ such choices.
If the units digit is $3$ or $5$, there are two choices for the hundreds digit (since we must exclude the units digit from the choices $2$, $3$, or $5$) and three choices for the tens digit (since we must exclude the units digit and the hundreds digit). Thus, there are $2 \cdot 3 \cdot 2 = 12$ such choices.
In all, there are $3 + 12 + 9 + 12 = 36$ odd numbers less than $600$ that can be formed if the digit $3$ is used at most once.
Case 2: The digit $3$ is used twice.
There must be at least two digits.
Two-digit numbers: There is just one, namely $33$.
Three-digit numbers: There are two three-digit odd numbers in which both the hundreds digit and tens digit are $3$'s, namely $335$ and $337$.
If both the the hundreds digit and units digit are $3$'s, there are four choices for the tens digit, namely $2$, $3$, $6$, or $7$.
If both the tens digit and units are $3$'s, the requirement that the number be less than $600$ means that the units digit is either $2$ or $5$.
Hence, there are $1 + 2 + 4 + 2 = 9$ odd numbers less than $600$ that can be formed if the digit $3$ is used twice.
Total: Combining the two mutually exclusive cases yields $36 + 9 = 45$ odd numbers less than $600$ that can be formed using the digits $2, 3, 3, 5, 6, 7$ without repetition.
Method 2: Tedious case work.
The requirement that the number must be odd means the units digit is either $3$, $5$, or $7$.
Case 1: Single-digit odd numbers.
There are three such numbers, namely $3$, $5$, and $7$.
Case 2: Two-digit odd numbers.
The units digit is $3$. Then we have five choices for the tens digit, namely $2$, $3$, $5$, $6$, or $7$.
The units digit is $5$. Then we have four choices for the tens digit, namely $2$, $3$, $6$, or $7$.
The units digit is $7$. Then we have four choices for the tens digit, namely $2$, $3$, $5$, or $6$.
Hence, it is possible to form $5 + 4 + 4 = 13$ odd two-digit numbers if we use the digits $2, 3, 3, 5, 6, 7$ at most once in each number.
Case 3: Three digit odd numbers less than $600$.
The units digit is $3$. Since the number must be less than $600$, the hundreds digit is either $2$, $3$, or $5$.
The units digit is $5$. Since the number must be less than $600$, the hundreds digit is either $2$ or $3$.
The units digit is $7$. Since the number must be less than $600$, the hundreds digit is either $2$, $3$, or $5$.
In all, the number of three-digit odd numbers less than $600$ that can be formed using the digits $2, 3, 3, 5, 6, 7$ is $$3 \cdot 4 + 3 + 4 + 2 \cdot 3 + 4 = 12 + 3 + 4 + 6 + 4 = 29$$
Total: Thus, we have a total of $3 + 13 + 29 = 45$ odd numbers less than $600$ that can be formed using the digits $2, 3, 3, 5, 6, 7$ without repetition.