Problem 42 - College Algebra

Question

In horse racing, a "trifecta" occurs when a bettor wins by selecting the fi st three fin shers in the exact order (1st place, 2 nd place, and 3 rd place \() .\) How many different trifectas are possible if there are 14 horses in a race?

Step-by-Step Solution

Answer

There are 2,184 different trifectas possible.

Step 1: Understand the Problem

We need to calculate how many different ways we can arrange 14 horses to finish in the top three positions: 1st, 2nd, and 3rd. This is essentially a permutation problem because the order of finishers matters.

Step 2: Calculate the Permutations

For the first position, there are 14 possible choices (any of the 14 horses can finish first). After a horse is chosen for the first position, there are 13 horses left for the second position. Finally, for the third position, there are 12 horses remaining. Therefore, the total number of different trifectas is calculated by the product: \[ 14 \times 13 \times 12 \]

Step 3: Perform the Calculation

Compute the expression from Step 2: \[ 14 \times 13 = 182 \] \[ 182 \times 12 = 2184 \] This gives us the total number of different trifectas possible.

Key Concepts

Combinatorial Problems

Combinatorial problems deal with counting the number of ways certain arrangements or selections can be made. These problems are essential in determining possibilities and tests how arrangements can vary in different scenarios. A common type of combinatorial problem focuses on permutations, where the order of selection or arrangement significantly matters. In the case of horse racing, if you're trying to predict the exact sequence of the first three horses finishing, understanding combinatorial problems helps you determine how many possible arrangements exist. This makes it easier to comprehend the complexity and the task's likelihood by breaking down all potential outcomes.

Factorial Calculation

Factorial calculation, denoted by the symbol \(!\), is fundamental in permutations and combinations. It involves multiplying a series of descending natural numbers. For instance, \(n!\) equals \(n \times (n-1) \times (n-2) \times \ldots \times 1\). This is crucial when you're calculating permutations as it represents the total ways to arrange \(n\) items. In our horse racing example, although we didn't explicitly calculate a factorial, the concept is in the background calculations of permutations where you list items in a specific order, like choosing 14 \(n\), 13 \(n-1\), and then 12 \(n-2\). While not directly written as \(14!\), partial factorial ideals, like \(14 \times 13 \times 12\), factor into the calculation, showcasing the logic behind such ordering problems.

Probability in Horse Racing

Probability in horse racing is vital for gamblers to estimate their chances of winning. When placing bets, probability helps determine the likelihood of events occurring, such as selecting the top three finishers in order. Calculating permutations, like in our exercise, aids in understanding the sheer number of possible outcomes, which indirectly provides insight into probability. For example, calculating 2,184 possible trifecta outcomes shows the complexity bettors face when choosing winners. The actual probability of selecting the correct trifecta can be computed by considering how many ways the bet can be structured versus the one winning outcome. This kind of understanding empowers bettors with the knowledge to make informed decisions.