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What would be the probability of putting item A between B and C, considering the different possible combinations (so BAC or CAB), out of 6 items in total.
So we need to calculate the probability for putting item A between B and C, with 3 other items and their possible combinations.
How would I go about calculating this, please?
We only care about the relative positions of A, B, and C. By symmetry, A will be in the middle position of these three letters in $1/3$ of the permutations.
If this seems too simple, consider arranging the letters A, B, C, D, E, F. Since they are distinct, this can be done in $6!$ ways. As for the favorable cases, we can place D in six ways, place E in five ways, and place F in four ways. Once we have done so, A must be placed in the middle open position. That leaves us with two ways to place B and one way to place C, which gives $6 \cdot 5 \cdot 4 \cdot 1 \cdot 2 \cdot 1$ favorable cases, so the probability that A is somewhere between B and C is $$\frac{6 \cdot 5 \cdot 4 \cdot 1 \cdot 2 \cdot 1}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = \frac{1}{3}$$ which confirms the symmetry argument given above.
If it were just AB there would be 2^2 combinations. [AA, AB, BA, BB] Since it's ABC, three are 3^3 combinations = 27. Out of these 27 you're wanting to know how many start and end with B_C or C_B. And have an "A" between. My non-mathmatically trained guess is 2 out of 27. BAC CAB are the only combinations that meet the criteria.
If you're asking about two sets [??? , ???], I think that would be 27^2 possibilities. But I'm guessing the 2/27 ratio would hold.
If the last letter of the first set can encompass an A with the 2nd letter of the second set : like ABC ABC, I'm wondering what could be squared.
Obviously, I'm guessing about all of this. But I tried.
New approach. For A to be between C and B, it must occupy position 2,3,4, or 5. (Position 1 and 6 would be missing a side). There are only 2 combinations at each one of these positions BAC or CAB. 5 X 2 = 10 possible combinations for each position. Now the last remaining calculation is how many combinations are possible with the remaining 3 letters. 3^3 = 27. 270 possible combinations out of 6^6 possible combinations = 270 / 46,656 = .005787. I'm just brainstorming here. I imagine a second calculation would be needed to weed out redundancies...
This is obviously so wrong. Any corrections in my approach are much appreciated.
