Question

20 cars are in a race. In how many different ways can cars finish in the top 3 positions?

Answer 1

We have 20 options of cars to finish in the first 3 positions.

From the context of the problem we know that:

• No car can be in two positions (No repetition)

,

• For a car to be first, is different from being in the second or third place (Order is important).

Then, we can conclude that this is a permutation of 20 in 3 without repetition and can be calculated as:

`\begin{gathered} P(n,r)=(n!)/((n-r)!) \\ P(20,3)=(20!)/((20-3)!)=(20!)/(17!)=20\cdot19\cdot18=6840 \end{gathered}`

Answer: 6840 ways.