Problem 9
Question
Wendy's Hamburgers once advertised that there were 256 different hamburgers available at Wendy's. This number was obtained by using the fundamental counting principle and considering whether to include each one of several different options on the burger. How many different optional items were used to get this number?
Step-by-Step Solution
Verified Answer
8 optional items.
1Step 1: Understanding the Counting Principle
The fundamental counting principle states that if there are several choices to make and the number of ways to make each choice is known, the total number of ways to make all choices is the product of the number of ways to make each choice.
2Step 2: Formulating the Problem
To find the number of optional items, denote the number of items by using the variable n. We then need to determine the number of possible combinations of including or excluding each optional item.
3Step 3: Setting Up the Equation
Each optional item has 2 choices: either it can be added to the burger or it cannot. Therefore, the number of different hamburgers can be represented by the equation \(2^n = 256\).
4Step 4: Solving for n Using Exponents
To find n, we need to solve the exponent equation \(2^n = 256\). We can rewrite 256 as a power of 2: \(256 = 2^8\). Therefore, \(n=8\).
5Step 5: Conclusion
By using the fundamental counting principle and solving the equation \(2^n = 256\), it is determined that there are 8 different optional items.
Key Concepts
ExponentiationCombinationsOptional Items
Exponentiation
Exponentiation is the mathematical operation where a number, known as the base, is multiplied by itself a certain number of times. The number of times it is multiplied is represented by an exponent.
For instance, in the equation \(2^3\), 2 is the base and 3 is the exponent. This means \(2^3 = 2 \times 2 \times 2 = 8\).
In the context of our Wendy's hamburgers problem, exponentiation is used to calculate the total number of combinations of optional items. Each item can either be included or not, which gives us two choices. If we have 'n' optional items, the total number of combinations is calculated as \(2^n\). Therefore, solving \(2^n = 256\) was crucial to find the number of optional items Wendy’s offers.
For instance, in the equation \(2^3\), 2 is the base and 3 is the exponent. This means \(2^3 = 2 \times 2 \times 2 = 8\).
In the context of our Wendy's hamburgers problem, exponentiation is used to calculate the total number of combinations of optional items. Each item can either be included or not, which gives us two choices. If we have 'n' optional items, the total number of combinations is calculated as \(2^n\). Therefore, solving \(2^n = 256\) was crucial to find the number of optional items Wendy’s offers.
Combinations
Combinations are a way to calculate how many different groups, or sets of items can be selected from a larger pool, where the order does not matter.
In our problem, we are not concerned with the order of the optional items on a hamburger, but rather, whether each item is included or not.
Because each item can either be included (1) or not included (0), each optional item has two possible states. Thus, with 'n' items, we use the combination formula \(2^n\) to determine the total number of combinations of optional items available for burgers at Wendy’s.
When we solved \(2^n = 256\), we found that there were exactly 8 optional items that customers could choose from.
In our problem, we are not concerned with the order of the optional items on a hamburger, but rather, whether each item is included or not.
Because each item can either be included (1) or not included (0), each optional item has two possible states. Thus, with 'n' items, we use the combination formula \(2^n\) to determine the total number of combinations of optional items available for burgers at Wendy’s.
When we solved \(2^n = 256\), we found that there were exactly 8 optional items that customers could choose from.
Optional Items
Optional items in the context of Wendy’s hamburgers refer to the extra toppings or ingredients that can be added to a burger. Each optional item provides a choice: to include it or to exclude it.
Examples of optional items might include cheese, lettuce, tomato, pickles, onions, ketchup, mustard, and bacon.
The key concept here is understanding how the availability of options adds to the complexity of choices. Each optional item effectively doubles the total number of different burgers you can create. Therefore, being able to include or exclude these items explains the use of the fundamental counting principle and exponentiation in solving this problem.
The final result showed us that with 8 optional items, Wendy’s could indeed offer 256 unique burgers.
Examples of optional items might include cheese, lettuce, tomato, pickles, onions, ketchup, mustard, and bacon.
The key concept here is understanding how the availability of options adds to the complexity of choices. Each optional item effectively doubles the total number of different burgers you can create. Therefore, being able to include or exclude these items explains the use of the fundamental counting principle and exponentiation in solving this problem.
The final result showed us that with 8 optional items, Wendy’s could indeed offer 256 unique burgers.
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