# What is the probability of putting an item between 2 other items

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.

• The only thing that matters here is the relative positions of objects A, B, and C. How many ways can they be arranged (ignoring the other objects)? How many of these ways place A between B and C (again ignoring the other objects)? Oct 3, 2019 at 17:51
• @N.F.Taussig Can I disregard the other 3 positions then? For example, if there are positions A,B,C,D,E,F and I just want to know about the possibilities for position A between B and C, can I disregard positions D,E, and F? And, is there a specific formula I can use to figure out the position of A between B and C? Oct 3, 2019 at 18:07

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.

• Be careful. The objects are distinct, so there are only two ways of arranging A and B and only $3! = 6$ ways of arranging A, B, and C. Oct 3, 2019 at 18:17

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.

• The problem does not say that A, B, and C are consecutive. If A is in the second position, you have to choose whether to place B or C in the first position. If you place B in the first position, you have four choices where to place C. Also, the remaining letters are distinct, so they can be arranged in the remaining three positions in $3!$ ways. Oct 3, 2019 at 19:13