Let's break down this problem.
We need to look for pairs of digits whose product is 24, and then arrange each pair into the maximum number of two-digit numbers.
The first part: let's find all of the factors of 24: 24 and 1, 12 and 2, 8 and 3, 6 and 4. Of these factors, only 8 and 3 and 6 and 4 are pairs where both numbers are single digit. This is an elementary number theory type of problem.
Now let's look at making each pair into a two-digit number. This is a combinatorics problem. The question really is: given two distinct things, how many ways can I arrange them in an order? If we look at 8 and 3 and pick a number to go first we can do it two ways: 8- and 3-. Now, if we picked the 8 to go first, we must pick the three to go next, and vice versa. By picking the first number, we have forced our choice of the second number. That means that for each pair of numbers we have two ways to make them into two-digit numbers.
Hence, two factor pairs, each of which can be arranged two ways, and we get the final answer of four such numbers. The numbers are 83, 38, 46, and 64.
If you want to push your understanding, why not try to do the same thing with 3 digits and 24? That is, how many three-digit numbers are there so that the product of their digits is 24? Now you have to factor into triplets instead of pairs and think about the different ways to order three numbers - and sometimes with duplicate numbers! (the final answer is nine, but can you prove it?)