Full solution

Q. There were

12

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

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Answer:

Identify the problem: Identify the problem. $\newline$ 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. $\newline$ The number of permutations of $n$ distinct objects taken $r$ at a time is given by $nPr = \frac{n!}{(n-r)!}$ . $\newline$ In this case, $n = 12$ (total students) and $r = 3$ (positions to fill).
Calculate the permutation: Calculate the permutation. $\newline$ Using the formula from Step $2$ , we calculate $12P3$ . $\newline$ $12P3 = \frac{12!}{(12-3)!}$ $\newline$ $= \frac{12!}{9!}$
Simplify the factorial expression: Simplify the factorial expression. $\newline$ We can simplify $12! / 9!$ by canceling out the common factorial terms. $\newline$ $12! / 9! = (12 \times 11 \times 10 \times 9!) / 9!$ $\newline$ $= 12 \times 11 \times 10$ $\newline$ $= 1320$
Verify the result: Verify the result. $\newline$ 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.