Problem 44 - College Algebra with Corequisite Support

Question

Hector wants to place billboard advertisements throughout the county for his new business. How many ways can Hector choose 15 neighborhoods to advertise in if there are 30 neighborhoods in the county?

Step-by-Step Solution

Answer

There are 155,117,520 ways to choose 15 neighborhoods out of 30.

Step 1: Understanding the Problem

Hector needs to choose 15 neighborhoods out of 30 available options. This is a combinatorial problem where the order of selection doesn't matter.

Step 2: Applying the Combination Formula

To find out how many different ways Hector can choose 15 neighborhoods out of 30, we use the combination formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]where \(n = 30\) and \(r = 15\).

Step 3: Calculating the Factorials

We calculate the necessary factorials: - \(30!\) is the factorial of 30.- \(15!\) is the factorial of 15.- \((30-15)! = 15!\) because 30 minus 15 equals 15.

Step 4: Substituting the Values

Substitute the factorials into the formula: \[ \binom{30}{15} = \frac{30!}{15! \times 15!} \]

Step 5: Simplifying the Expression

The expression simplifies by canceling out the common \(15!\) in both the numerator and the denominator, to avoid directly calculating very large numbers:\[ \binom{30}{15} = \frac{30 \times 29 \times 28 \times \ldots \times 16}{15 \times 14 \times 13 \times \ldots \times 1} \]

Step 6: Final Calculation

Perform the calculation step-by-step or using a calculator for the simplified expression: \[ \binom{30}{15} = 155117520 \]

Key Concepts

Combination Formula

When solving problems like choosing neighborhoods for advertisements, we often use the combination formula. The combination formula helps us count the number of ways to select a subset from a larger set, where the order doesn't matter. It is denoted by \( \binom{n}{r} \) and calculated with the formula:\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]Here, \(n\) represents the total number of items, and \(r\) indicates how many we want to select. In Hector's problem, he chooses 15 neighborhoods from 30.The key feature of combinations is that the order of selection does not matter. This is different from permutations, where the order does matter. So, whenever you see a problem where you are simply picking items and the order doesn't matter, think combinations!
To solve such problems efficiently, it's crucial to apply this formula correctly. Ensure you identify the total items and how many you need to pick. Substituting these into the formula gives you the number of combinations, guiding you to the solution.

Factorials

The concept of factorials is vital in combinatorics, especially when dealing with the combination formula. Factorial, denoted by an exclamation mark (\(!\)), is the product of all positive integers up to a certain number. For instance, \(5!\) (read as "five factorial") equals \(5 \times 4 \times 3 \times 2 \times 1 = 120\).
Factorials grow quickly into large numbers. For example, \(30!\) involves multiplying together every integer from 1 to 30. Due to its size, calculating factorials directly can be cumbersome. In our exercise, we wisely avoid calculating \(30!\) altogether by using the properties of factorial division in the combination formula:

• Instead of calculating \(30!\) completely, we calculate \(30 \times 29 \times \ldots \times 16\), because the rest from 1 to 15 cancels out with \(15!\) in the denominator.
• Similarly, \(15!\) simplifies since it's in both the numerator and denominator of our fraction.

This not only simplifies calculations but also makes the task more manageable with calculators or mental math. Learning to manipulate factorials efficiently can save time and effort in combinatorial problems.

Combinatorial Problem

A combinatorial problem involves finding the number of ways to arrange or select items according to certain rules. There are two main types: combinations and permutations. In combinations, like Hector's billboard task, the order doesn't matter. This distinguishes it from permutations where the sequence of choice is essential.
To tackle these problems, it’s crucial to:

• Identify whether the solution requires permutations or combinations.
• Use the relevant formula based on the problem’s needs.
• Simplify calculations whenever possible to make the process smoother.

Hector’s problem exemplifies a real-world combinatorial task. It asks us to determine how many different sets of neighborhoods he can select. In practice, recognizing when a problem is combinatorial helps apply systematic approaches to count efficiently.
The beauty of combinatorial problems is that they appear in diverse scenarios, from simple games to complex logistical tasks. Understanding these principles allows you to solve a broad spectrum of counting problems with confidence and ease.