Question 1: Bella Works At Sub Shop And Wants To Know How Ma

Question 1bella Works At Sub Shop And Wants To Know How Many Differen

Question #1 Bella works at sub shop and wants to know how many different types of subs she can make considering all of the available options. There are six types of rolls, five types of meats, eight types of vegetables, and three types of cheeses. If two choices are made from each of the four categories, how many types of subs can Bella make?

Question #2 How many odd, five-digit numbers can be created from the digits 1 to 5 if repetition is allowed?

Question #3 How many ways can four calculators be chosen for testing from a group of 10?

Paper For Above instruction

This paper addresses three combinatorial problems involving arrangements and selections from various categories. The approach to solving each problem involves applying principles of counting, permutations, and combinations. The goal is to systematically analyze each problem, demonstrate the calculation process, and interpret the results within the context of each question.

Problem 1: Number of Different Sub Types at a Sub Shop

Bella's sub shop offers a variety of options across four categories: rolls, meats, vegetables, and cheeses. The problem specifies that two choices are made from each category, with limited options: six types of rolls, five types of meats, eight types of vegetables, and three types of cheeses. The question is: How many unique subs can Bella create when selecting two options per category?

To approach this problem, we recognize that the selection process involves choosing 2 options from each category, possibly with repetitions, assuming the store allows double portions or repeated ingredients. If repetition is permitted, then the number of combinations with repetition for each category can be calculated using the formula:

n + r - 1 choose r, where n is the number of options in a category, and r is the number of choices made.

For each category:

  • Rolls: n = 6, r = 2 → C(6 + 2 - 1, 2) = C(7, 2) = 21
  • Meats: n = 5, r = 2 → C(5 + 2 - 1, 2) = C(6, 2) = 15
  • Vegetables: n = 8, r = 2 → C(8 + 2 - 1, 2) = C(9, 2) = 36
  • Cheeses: n = 3, r = 2 → C(3 + 2 - 1, 2) = C(4, 2) = 6

Next, we multiply the number of choices across categories to find the total number of possible unique subs:

Total options = 21 (rolls) × 15 (meats) × 36 (vegetables) × 6 (cheeses) = 21×15×36×6

Calculating step-by-step:

  • 21 × 15 = 315
  • 315 × 36 = 11,340
  • 11,340 × 6 = 68,040

Therefore, Bella can make 68,040 different types of subs considering all options.

Note: If the problem intended that two different choices are made without repetition (i.e., choosing two distinct ingredients from each category), then the calculation would change to combinations without repetition. For simplicity, and aligning with typical menu choices, the assumption of repetition is accepted here.

Problem 2: Number of Odd, Five-Digit Numbers from Digits 1 to 5

In this problem, we need to determine how many five-digit numbers can be formed using digits 1 through 5, allowing repetition, such that the resulting number is odd.

The key here is that for a number to be odd, its units (ones) digit must be odd. From the given digits, the odd digits are 1, 3, and 5, totaling 3 options for the units digit.

For each position:

  • The units digit: 3 options (1, 3, or 5)
  • The remaining four digits: each can be any of the five digits (1-5), with repetition allowed, so 5 options each

The total number of five-digit odd numbers is thus:

Number of choices for each position = 5 (for first four positions) × 3 (for units digit)

Calculating:

  • First digit: 5 options
  • Second digit: 5 options
  • Third digit: 5 options
  • Fourth digit: 5 options
  • Fifth digit (units): 3 options

Total numbers = 5 × 5 × 5 × 5 × 3 = 5^4 × 3 = 625 × 3 = 1,875

Thus, there are 1,875 such five-digit odd numbers.

Problem 3: Choosing 4 Calculators from 10

This problem involves calculating the number of ways to select 4 calculators from a group of 10 for testing, where order does not matter.

Since the order of selection is irrelevant, this is a combination problem. The formula for combinations is:

n choose r = n! / [r! (n - r)!]

Applying the formula:

C(10, 4) = 10! / [4! × (10 - 4)!] = 10! / (4! × 6!)

Calculations:

10! = 3,628,800

4! = 24

6! = 720

Thus, C(10, 4) = 3,628,800 / (24 × 720) = 3,628,800 / 17,280 = 210

Therefore, there are 210 ways to choose four calculators from the group of ten.

Conclusion

In summary, the analysis of these three problems demonstrates the application of fundamental combinatorial principles. For the sub shop, considering choices with repetition yields a total of 68,040 unique subs. For forming five-digit odd numbers, the calculations show there are 1,875 possibilities. Lastly, selecting four calculators from ten results in 210 combinations, illustrating the use of the combination formula. These problems underscore how mathematical principles can be effectively employed to solve real-world and theoretical problems, highlighting the importance of understanding permutations and combinations in various contexts.

References

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