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There were 18 students running in a race. How many different arrangements of first, second, and third place are possible?
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There were 1818 students running in a race. How many different arrangements of first, second, and third place are possible?\newlineAnswer:

Full solution

Q. There were 1818 students running in a race. How many different arrangements of first, second, and third place are possible?\newlineAnswer:
  1. Calculate Permutations Formula: To determine the number of different arrangements for the first three places, we need to calculate the permutations of 1818 students taken 33 at a time. The formula for permutations is P(n,k)=n!(nk)!P(n, k) = \frac{n!}{(n - k)!}, where nn is the total number of items and kk is the number of items to arrange.
  2. Calculate Factorial of 1818: First, we calculate the factorial of 1818, which is 18!=18×17×16××118! = 18 \times 17 \times 16 \times \ldots \times 1.
  3. Calculate Factorial of Difference: Next, we calculate the factorial of the difference between the total number of students and the number of places, which is (183)!=15!=15×14××1(18 - 3)! = 15! = 15 \times 14 \times \ldots \times 1.
  4. Use Permutation Formula: Now, we use the permutation formula P(18,3)=18!(183)!=18!15!P(18, 3) = \frac{18!}{(18 - 3)!} = \frac{18!}{15!}.
  5. Simplify Expression: We simplify the expression by canceling out the common factorial terms: P(18,3)=18×17×161=18×17×16P(18, 3) = \frac{18 \times 17 \times 16}{1} = 18 \times 17 \times 16.
  6. Perform Multiplication: Finally, we perform the multiplication: 18×17×16=489618 \times 17 \times 16 = 4896.

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