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

User Maddouri
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1 Answer

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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.

User Paceman
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