To solve this, let's first consider the total number of combinations when there are no restrictions. Since there are 4 friends, there are 4! (4 factorial) ways to arrange them, which equals 24 combinations.

Now, let's count the number of combinations where A and C are next to each other. Since A and C can be arranged in 2 ways, and for each arrangement, the other two friends (B and D) can be arranged in 2! ways, the total number of combinations where A and C are next to each other is 2 * 2! = 4.

So, the number of combinations where A and C cannot be next to each other is the total number of combinations minus the number of combinations where A and C are next to each other:

24 - 4 = 20 combinations.

[–] Umbrias 2 points 7 months ago

You can have it try again over and over, even while telling it the answer is 12, and it hallucinates basically random numbers to boot.