Assume that there are nine parking spaces next to one another in a
parking lot.  Nine cars need to be parked by an attendant.  Three of
the cars are expensive sports cars, three are large domestic cars and
three are compacts.  Assuming the attendant parks the cars at random,
what is the probability that the sports cars are parked next to each
other?

Good evening,

First off, I assume the parking spots can be numbered 1 through 9 such
that each adjacent number represents an adjacent parking spot.

I'll start with the "logical" explanation, and then follow it up with
the actual math.  This is a probability problem that we'll solve by
counting.  First, we need to count the total number of ways that these
9 cars can be arranged in the 9 parking spaces.  Next, we count how
many of those orders have the sports cars next to each other.  Since
your question states that the probability of each possible order
occuring is the same, we can simply divide the number of ways the
sports cars can end up together by the total number of ways the cars
can be arranged, and this gives the probability of the sports cars
ending up together.

The first part of this is easy.  For parking space 1, any of the 9
cars can be parked.  In the second spot, only 8 cars can (the original
9, minus the one in spot 1).  In the third spot, 7 cars can be parked
(the original 9, minus the car is spot 1, minus the car in spot 2),
and so on.  The 9 cars can be parked in 9! = 9 * 8 * 7 * . . . * 2 * 1
unique ways.

If the sports cars are to be together, they must be in consecutive
parking spaces.  For example {1, 2, 3} or {6, 7, 8}.  There are 7 such
consecutive runs of 3 stalls.  For each of these runs, the sports cars
can be in several orders (as above, 3! different ways to order 3 cars
in 3 parking stalls).  Furthermore, once parking stalls are chosen for
the sports cars, and the sports cars are ordered within those stalls,
the other cars must be considered.  For each run, and each order of
sports cars, the remaining 6 cars can be placed in 6! different
orders.

Now we have 9! total orders that the cars can be in, and 7 * 3! * 6!
of them have the three sports cars next to each other.

(7 * 3! * 6!) / 9! = (3! * 7!) / 9!

(3 * 7!) / 9!  = 3 / (9 * 8)

3 / (9 * 8) = 1 / 24, or just over 4%.

You picked a good time to ask this question - I'm taking a probability
course myself, and we just covered this material.  If you need it
explained differently, just ask and I'll show you a different way to
attack the problem.

-Haversian

All the reasoning is right--just one math error.

(7 * 3! * 6!) / 9! = (3! * 7!) / 9!

(3 * 7!) / 9!  = 3 / (9 * 8)

3 / (9 * 8) = 1 / 24, or just over 4%.

In going from the first line to the second line above, a 3! got
changed to a 3, so the answer is off by a factor of 2.  The correct
answer is 1/12, or about 8%.  Another way to see this is that there
are (9 choose 3) ways to pick 3 parking spaces out of 9.  That's
9!/3!6! = 84 ways.  As Haversian pointed out, 7 of these ways have all
three spaces together, so the probability is 7/84 = 1/12.