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Chapter 6: Problem 23

In a state lottery, there are 15 finalists eligible for the Big Money Draw. In how many ways can the first, second, and third prizes be awarded if no ticket holder can win more than one prize?

Short Answer

Expert verified
There are 2730 ways to award the first, second, and third prizes among the 15 finalists if no ticket holder can win more than one prize. This is calculated using the permutation formula: P(15, 3) = 15! / 12! = 15 * 14 * 13 = 2730.

Step by step solution

01

Identify the problem type

This is a permutation problem because we care about the order of the prizes (first, second, and third).
02

Use the Permutation Formula

The formula for permutations is: P(n, r) = n! / (n - r)! Where n is the number of elements, r is the number of elements we are choosing, and ! denotes a factorial.
03

Plug in the values into the Permutation Formula

We need to find the permutations of 15 finalists (n) taken 3 at a time (r). P(15, 3) = 15! / (15 - 3)!
04

Simplify the factorial expression

Now, let's simplify the expression. P(15, 3) = 15! / 12!
05

Compute the answer

To compute the answer, we can cancel out the common 12! factor in both the numerator and the denominator: P(15, 3) = (15 * 14 * 13 * 12!) / 12! P(15, 3) = 15 * 14 * 13 Now, let's multiply these numbers to get the final answer: 15 * 14 * 13 = 2730 So, there are 2730 ways to award the first, second, and third prizes among the 15 finalists if no ticket holder can win more than one prize.

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