Six people, designated as A, B, C, D, E, and F, are arranged in linear order.
Assuming each possible order is equally likely, find the probability of the
following events:
a) there is exactly one person between A and B,
A _ B _ _ _
_ A _ B _ _
_ _ A _ B _
_ _ _ A _ B
B _ A _ _ _
_ B _ A _ _
_ _ B _ A _
_ _ _ B _ A
In each of the above 8 ways 1 person is between A and B, there are 4!
ways to seat C, D, E, and F.
That's 8·4! = 8·24 = 192 ways.
There are 6! = 720 ways they could sit in any order so the probability
of there being one person between them is 192 ways out of 720, or
which reduces to
b) there are exactly two people between A and B,
A _ _ B _ _
_ A _ _ B _
_ _ A _ _ B
B _ _ A _ _
_ B _ _ A _
_ _ B _ _ A
In each of the above 6 ways 2 people are between A and B, there are 4!
ways to seat C, D, E, and F.
That's 6·4! = 6·24 = 144 ways.
There are 6! = 720 ways they could sit in any order so the probability
of there being one person between them is 144 ways out of 720, or
which reduces to
c) there are four people between A and B.
A _ _ _ _ B
B _ _ _ _ A
In each of the above 2 ways 4 people are between A and B, there are 4!
ways to seat C, D, E, and F.
That's 2·4! = 2·24 = 48 ways.
There are 6! = 720 ways they could sit in any order so the probability
of there being one person between them is 48 ways out of 720, or
which reduces to
Edwin