6 children arrange themselves at random in a circle.?

Question

What is the probability that the tallest and shortest are together? My own solution is [({6-2+1}-1)!2!]/[6-1]! = 48/120 = 2/5. But I'm not sure whether that's the solution. I know that number of ways of arranging n people in a circle is (n - 1)!.. But here I have 2 people who are supposed to be next to each other ie the tallest and shortest. In my numerator, i took 6-2 to remove the 2 people, add 1 to add in the 2 people who are now being considered as 1 group, -1 again cos of formula for permutation in a circle, *2! to permutate the people in the group. is that correct or do I have to apply the circle formula and take (2-1)! ? My instincts says the "2" doesnt have to be subtracted by 1 but can someone help me put it in a clearer perspective as to why we don't have to? And why do we have to take (n-1)! for permutations in a circle?

Answer 1

Probability that the shortest person is in the circle: 1

When we know that shortest person is at position i then probability of tallest at position i-1 or i+1 is 2/5.

So probability that both are next to eachother is 2/5.

Answer 2

Common sense ans is 2/5

why say tallest or shortest has a spot. to be together the other must be to the right or to the left. THere is equal chance of that so that since there are five other spots for the other to sit but only two that has them together its 2/5 as your fancy pancy math I dont understand so states

Answer 3

i think the answer is (6-1)!*2
(1)first you consider the tallest and the shortesh as one guy and you find how many possibilities is there for aranging 1 guy in a circle of 5 (including him) = 5! (6-1)*
(2)but this "guy" is two guys that can stand in this way : short-tall, or: tall-short. => for every posibilty in (1) there are two ways.
therefore: (6-1)!*2

Answer 4

Start from tallest (call him A, call shortest F).
Who is sitting next to A? (no regard to direction)
B,C
B,D
B,E
B,F
C,D
C,E
C,F
D,E
D,F
E,F

10 pairings, of which 4 contain F