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Sagot :
We have 30 cars in a race and we have to find the possible ways are for the top 3 positions.
This is a permutation, as the order does matter. It is also without repetition, as each car is unique.
Then, we have a permutation without repetition of 30 elements in 3 places.
We can express it as:
[tex]\begin{gathered} P(n,r)=\frac{n!}{(n-r)!} \\ P(30,3)=\frac{30!}{(30-3)!}=\frac{30!}{27!}=30\cdot29\cdot28=24360 \end{gathered}[/tex]NOTE: we could have also deducted this number in another way.
We have 30 cars that can be in the first place. Then, for the second place we are left with 29 options, as one car is already in the first place. Finally, for the third place we have 28 options.
If we multiply the options we have 30*29*28 = 24360.
Answer: 24,360 ways.
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