Six people, including Alice, Bob, and Connie, form a queue in random order. What is the probability when Bob is between Alice and Connie?

Question

I figured out several patterns of queues like this:

$$ABC•••$$ $$A•BC••$$ $$AB•C••$$ $$AB••C•$$ $$AB•••C$$

Am I coming through the right way or not? Can anyone give me some hints, please?

Answer 1

Hint: the positions of the other three people are irrelevant. Just look at the permutations of $(A,B,C)$.

Answer 2

When you say "between," it is unclear whether you mean directly between, or whether A...B.C counts as B being between A and C. If this is the case, there is a 1/3 chance any one of the three is "between" the others.

Answer 3

• If you assume that $A,B,C$ don't have to be next to each other then the other people don't matter. You would have a total of $3!$ ways of organizing $A,B,C$ in line. How many out of those permutations have $B$ in the middle?

• If you assume that $A,B,C$ have to be next to each other: make $ABC$ one block. You would thus have $4$ objects: $ABC, D, E, F$. Here, $D,E,F$ correspond to the other people in line. Thus, you would have $4!$ permutations where $B$ is between $A$ and $C$. However, since a queue has an order, you also have to consider the block $CBA$, which gives $4!$ additional ways to obtain the desired result. This gives us a total of $2 \times 4! = 48$ ways where $B$ is directly in between $A$ and $C$. Dividing amongst the $6!$ possible ways to organize the people in line gives you the desired result.

Answer 4

First we need to select 3 places out of 6. We can do this in $6\choose 3$ ways. We know that Bob is always going to occupy the middle place (out of the selected three). The other two places can be occupied by Alice and Conie in $2!$ ways. So the total number of arrangements become ${6\choose 3}.2!$
But wait! We need to place the other three people as well. This can be done in $3!$ ways
Hence the no. of arrangements becomes ${6\choose 3}.2!.3!=240$. Since the total no. of possible arrangements is $6!=720$, the probability of desired event is $\frac{240}{720}=\frac13$