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There were 20 students running in a race. How many different arrangements of first, second, and third place are possible?
Answer:

There were $20$ students running in a race. How many different arrangements of first, second, and third place are possible? $\newline$ Answer:

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Experimental probability

Full solution

Q. There were

20

students running in a race. How many different arrangements of first, second, and third place are possible?

\newline

Answer:

Calculate Permutations Formula: To determine the number of different arrangements for the first three places, we need to calculate the number of permutations of $20$ students taken $3$ at a time. The formula for permutations is $P(n, k) = \frac{n!}{(n - k)!}$ , where $n$ is the total number of items and $k$ is the number of items to arrange.
Calculate Factorial of $20$ : First, we calculate the factorial of $20$ , which is $20! = 20 \times 19 \times 18 \times \ldots \times 1$ .
Calculate Factorial of $17$ : Next, we calculate the factorial of the difference between the total number of students and the number of places, which is $(20 - 3)! = 17! = 17 \times 16 \times 15 \times \ldots \times 1$ .
Use Permutation Formula: Now, we use the permutation formula $P(20, 3) = \frac{20!}{(20 - 3)!} = \frac{20!}{17!}$ .
Simplify Expression: We simplify the expression by canceling out the common factorial terms: $P(20, 3) = \frac{20 \times 19 \times 18}{1} = 20 \times 19 \times 18$ .
Perform Multiplication: We perform the multiplication: $20 \times 19 \times 18 = 6840$ .
Final Result: Therefore, there are $6840$ different arrangements of first, second, and third place possible among $20$ students running in a race.

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