1. ## Probability

There are 9 cars and 9 parking spaces in a row. If 3 sport cars are randomly in the group of 9 cars, what is the probability that the sports cars are parked in a row?

For this question is found the number of ways the sports cars can be parked in a row in 9 spots. They can be parked in 7*6 different ways out of a possible 9!.. correct?

Doesnt this seem a little low?

2. Originally Posted by mathlete2
There are 9 cars and 9 parking spaces in a row. If 3 sport cars are randomly in the group of 9 cars, what is the probability that the sports cars are parked in a row?

For this question is found the number of ways the sports cars can be parked in a row in 9 spots. They can be parked in 7*6 different ways out of a possible 9!.. correct?

Doesnt this seem a little low?
for each of the 42 ways of parking the 3 sports cars there are 6! arrangements of the remaining cars, so the required probability is:

(42 6!)/9! ~= 0.0833

RonL

3. Hello, mathlete2!

Another approach . . .

There are 9 cars and 9 parking spaces in a row.
If 3 sport cars are randomly in the group of 9 cars,
what is the probability that the sports cars are adjacent?

There are: . $9!$ possible arrangments.

Duct-tape the 3 sports cars together.
. . Then there are 7 "cars" to arrange: . $\boxed{SSS}, X, X, X, X, X, X$
There are: . $7!$ permutations.

But the 3 sports cars can be permuted in $3!$ ways.

Hence, there are: . $(7!)(3!)$ ways for the sports cars to be adjacent.

Therefore, the probability is: . $\frac{(7!)(3!)}{9!} \;=\;\frac{1}{12}$