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:
The difference between 3.5 and 4 is substantial. Here is what 4 says
To find the number of combinations in which four friends (A, B, C, and D) can stand in line such that A and C are not next to each other, we can use the following approach:
Total Combinations: First, calculate the total number of ways in which four people can be arranged in a line. This can be calculated by (4!) (4 factorial), since there are 4 slots to fill, each choice reducing the number of available choices by one for the next slot.
[
4! = 4 \times 3 \times 2 \times 1 = 24
]
Unwanted Combinations (Where A and C are next to each other):
Consider A and C as a single unit. This effectively reduces the number of units to arrange from 4 to 3 (the AC unit, B, and D).
These three units can be arranged in (3!) ways:
[
3! = 3 \times 2 \times 1 = 6
]
However, within the AC unit, A and C can switch places. So, there are 2 ways to arrange A and C within their unit.
Therefore, the total number of arrangements where A and C are next to each other is:
[
3! \times 2 = 6 \times 2 = 12
]
Subtracting Unwanted Combinations: Subtract the number of unwanted combinations (where A and C are next to each other) from the total number of combinations:
[
24 - 12 = 12
]
Thus, there are 12 combinations where A and C are not next to each other.
Logic. As an example, non textbook math questions. I asked ChatGPT 3.5 this:
It answered 20, the correct answer is 12.
All possible conbinations
abcd,abdc,adbc,adcb, cbad,cbda,cdba,cdab, bcda,badc, dcba,dabc
Its answer
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.
The difference between 3.5 and 4 is substantial. Here is what 4 says
Total Combinations: First, calculate the total number of ways in which four people can be arranged in a line. This can be calculated by (4!) (4 factorial), since there are 4 slots to fill, each choice reducing the number of available choices by one for the next slot. [ 4! = 4 \times 3 \times 2 \times 1 = 24 ]
Unwanted Combinations (Where A and C are next to each other):
Subtracting Unwanted Combinations: Subtract the number of unwanted combinations (where A and C are next to each other) from the total number of combinations: [ 24 - 12 = 12 ]
Thus, there are 12 combinations where A and C are not next to each other.