In how many different ways can four students stand in a straight line if two of the students refuse to stand next to each other?
This problem involves counting the number of possible arrangements of four students, which is a type of permutation. However, not all permutations are allowed, since two of the students have a restriction on their positions. Let's label the four students as A, B, C, and D, and assume that A and B are the ones who do not want to stand next to each other. One way to approach this problem is to consider the opposite scenario: how many ways can A and B stand next to each other? If we treat A and B as a single unit, then we have three units to arrange: AB, C, and D. The number of ways to arrange three units in a line is 3!, or 6. However, within the unit AB, we can also swap the positions of A and B, so we need to multiply by 2. Therefore, the number of ways that A and B can stand next to each other is 3! * 2, or 12. Now, we can use the complement principle to find the number of ways that A and B do not stand next to each other. Since the total number of ways to arrange four students in a line is 4!, or 24, we can subtract the number of ways that A and B stand next to each other from the total. So, the number of ways that A and B do not stand next to each other is 4! - 3! * 2, or 24 - 12, or 12. 