There were 12 students running in a race. How many different arrangements of first, second, and third place are possible? Answer:

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Identify the problem: Identify the problem. We need to find the number of different ways to arrange the first three places in a race with 12 students. This is a permutation problem because the order of the winners matters.Determine the formula: Determine the formula for permutations. The number of permutations of n distinct objects taken r at a time is given by nPr = n! / (n-r)!. In this case, n = 12 (total students) and r = 3 (positions to fill).Calculate the permutation: Calculate the permutation. Using the formula from Step 2, we calculate 12P3. 12P3 = 12! / (12-3)! = 12! / 9!Simplify the factorial expression: Simplify the factorial expression. We can simplify 12! / 9! by canceling out the common factorial terms. 12! / 9! = (12 × 11 × 10 × 9!) / 9! = 12 × 11 × 10 = 1320Verify the result: Verify the result. We have calculated that there are 1320 different arrangements for the first three places among 12 students. This makes sense because as we select each position, we have fewer students to choose from for the next position.

There were $12$ students running in a race. How many different arrangements of first, second, and third place are possible? $\newline$ Answer:

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