Problem 38

To solve this problem, we analyze each of the given options step by step:

- **Option A: Forming a four-digit number without repeating digits using $1$, $2$, $3$, and $5$**

We can choose any of the four digits for each place in the four-digit number. Thus, the total number of ways to form such a number is given by the permutation formula ${A}_{4}^{4}$, which calculates the number of ways to arrange $4$ items in $4$ positions, without repetition. This calculation is as follows:
$${A}_{4}^{4} = 4 \times 3 \times 2 \times 1 = 24$$
Therefore, $24$ numbers can be formed, making option A correct.

- **Option B: Forming $18$ odd numbers**

To form an odd number, the units digit must be $1$, $3$, or $5$. There are $3$ possibilities for the units digit. For the remaining three positions (thousands, hundreds, and tens), we can arrange the remaining $3$ digits in $3{A}_{3}^{3}$ ways, which is calculated as:
$$3{A}_{3}^{3} = 3 \times (3 \times 2 \times 1) = 18$$
Hence, $18$ odd numbers can be formed, making option B correct.

- **Option C: Forming $10$ even numbers**

To form an even number, the units digit must be $2$ (since it's the only even number among the given digits). There is $1$ possibility for the units digit. For the remaining three positions, we can arrange the remaining $3$ digits in ${A}_{3}^{3}$ ways, which is:
$${A}_{3}^{3} = 3 \times 2 \times 1 = 6$$
Therefore, only $6$ even numbers can be formed, making option C incorrect.

- **Option D: Forming numbers greater than $2000$**

To form a four-digit number greater than $2000$, the thousands digit must be $2$, $3$, or $5$ (not $4$ as mentioned in the standard solution, which seems to be a mistake since $4$ is not an option). There are $3$ possibilities for the thousands digit. For the remaining three positions, we can arrange the remaining $3$ digits in $3{A}_{3}^{3}$ ways, which is:
$$3{A}_{3}^{3} = 3 \times (3 \times 2 \times 1) = 18$$
Thus, $18$ numbers greater than $2000$ can be formed, making option D correct.

Given the analysis above, the correct options are A, B, and D. Therefore, the final answer is encapsulated as:
$$\boxed{\text{A, B, and D}}$$