A group of twelve people are going to take a ride on a roller coaster at
the same time in two different trains, one of which cannot fit more than
8 passengers, and the other cannot fit more than 7. In how many ways
can the group take a ride?
To solve this problem, we can use combinatorics.
First, let's consider the train that can fit 8 passengers. We need to determine the number of ways to choose 8 people out of the 12 to ride in this train.
We can calculate this using the formula for combinations:
C(n, r) = n! / ((n-r)! * r!)
where C(n, r) represents the number of ways to choose r items from a set of n items without considering the order.
In this case, n = 12 (total number of people) and r = 8 (number of people who can ride in the first train).
C(12, 8) = 12! / ((12-8)! * 8!)
= 12! / (4! * 8!)
= (12 * 11 * 10 * 9) / (4 * 3 * 2 * 1)
= 495
So, there are 495 ways to select 8 people from the group of 12 to ride in the first train.
Now, let's consider the second train that can fit 7 passengers. We need to determine the number of ways to choose 7 people out of the remaining 4 people who are not riding in the first train.
Using the same formula for combinations:
C(4, 7) = 4! / ((4-7)! * 7!)
= 4! / (-3! * 7!)
= 0 (since negative factorials are not defined)
Since we cannot choose 7 people from only 4, the number of ways for the group to take a ride is 0.
Therefore, the total number of ways for the group of 12 people to take a ride in the roller coaster is 495 * 0 = 0.