There are $9$ athletes at a track meet. How many different ways can they finish first, second, and third? $\newline$ Answer:

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Q. There are

9

athletes at a track meet. How many different ways can they finish first, second, and third?

\newline

Answer:

Calculate Permutations: To determine the number of different ways $9$ athletes can finish first, second, and third, we need to calculate the permutations of $9$ athletes taken $3$ at a time. This is because the order in which they finish matters.
Apply Permutation Formula: The formula for permutations of $n$ items taken $r$ at a time is $nPr = \frac{n!}{(n-r)!}$ , where $\text{“!“}$ denotes factorial, which is the product of all positive integers up to that number.
Calculate Factorials: First, we calculate the factorial of $9$ , which is $9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ .
Simplify Fraction: Now, we calculate the factorial of $(9-3)$ , which is $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ .
Multiply Factors: Using the permutation formula, we find the number of ways $9$ athletes can finish first, second, and third as $9P3 = \frac{9!}{(9-3)!} = \frac{9!}{6!}$ .
Final Result: Substituting the values we calculated for $9!$ and $6!$ , we get $9P3 = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$ .
Final Result: Substituting the values we calculated for $9!$ and $6!$ , we get $9P3 = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$ .We can simplify the fraction by canceling out the common factors in the numerator and the denominator. The $6 \times 5 \times 4 \times 3 \times 2 \times 1$ in the denominator cancels out with the same factors in the numerator, leaving us with $9P3 = 9 \times 8 \times 7$ .
Final Result: Substituting the values we calculated for $9!$ and $6!$ , we get $9P3 = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$ .We can simplify the fraction by canceling out the common factors in the numerator and the denominator. The $6 \times 5 \times 4 \times 3 \times 2 \times 1$ in the denominator cancels out with the same factors in the numerator, leaving us with $9P3 = 9 \times 8 \times 7$ .Multiplying the remaining factors, we get $9P3 = 9 \times 8 \times 7 = 72 \times 7 = 504$ .
Final Result: Substituting the values we calculated for $9!$ and $6!$ , we get $9P3 = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$ .We can simplify the fraction by canceling out the common factors in the numerator and the denominator. The $6 \times 5 \times 4 \times 3 \times 2 \times 1$ in the denominator cancels out with the same factors in the numerator, leaving us with $9P3 = 9 \times 8 \times 7$ .Multiplying the remaining factors, we get $9P3 = 9 \times 8 \times 7 = 72 \times 7 = 504$ .Therefore, there are $504$ different ways for $9$ athletes to finish first, second, and third at a track meet.