In a car race, there are 6 chevrolets, 4 fords, and 2 pontiacs. In how many ways can the cars finish if we consider only the makes of the cars?
12!/6!4!2! = 31933440
To determine the number of ways the cars can finish, we need to calculate the number of possible permutations for the makes of the cars.
First, we add up the total number of cars: 6 Chevrolets + 4 Fords + 2 Pontiacs = 12 cars.
Now, we need to think about how many ways we can arrange these 12 cars. This can be done using permutations.
In this case, because we are only considering the makes of the cars, we can think of each make as a distinct item. Therefore, the number of ways to arrange the cars is the same as arranging these 12 distinct items.
To find the number of permutations, we can use the formula for permutations of n distinct items, which is n! (n factorial).
For this situation, the number of permutations is 12! (12 factorial).
Calculating 12!:
12! = 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
After performing the multiplication, we get:
12! = 479,001,600
So, there are 479,001,600 different ways the cars can finish based on their makes.
Therefore, in a car race with 6 Chevrolets, 4 Fords, and 2 Pontiacs, there are 479,001,600 possible ways for the cars to finish.