Namemath125 Unit 8 Submission Assignment Answer Form Countin ✓ Solved

Namemath125 Unit 8 Submission Assignment Answer Formcounting Techniq

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Introduction

Counting techniques, including permutations and combinations, are fundamental concepts in combinatorics, used to determine the number of ways objects can be arranged or selected. Understanding the differences between these methods and their respective formulas is essential for solving various real-world problems involving probability, organization, and arrangement.

Part A: Combinations and Permutations

Question 1: Differences between permutations and combinations

Permutations and combinations are both methods for counting arrangements or selections of objects, but they differ primarily in whether the order of objects matters. In permutations, order is significant; for example, arranging books on a shelf. In combinations, order is irrelevant; selecting team members from a group is an example where only the group composition matters, not the order in which they are selected.

The formula for permutations of n objects taken r at a time is:

P(n, r) = n! / (n - r)!

The formula for combinations of n objects taken r at a time is:

C(n, r) = n! / [r! * (n - r)!]

Question 2: License plate format and calculating possibilities

My State’s Name: California

Sample License Plate Format: 1A2B3C4D5E6F7

Character Rule: Each license plate consists of seven characters, each either a number (0-9) or a capital letter (A-Z). No character repeats within the same license plate.

Number of unique license plates (no duplication of characters within each plate):

• The selection involves seven distinct characters out of 36 possible (10 digits + 26 letters).
• This is a permutation problem because order matters (the sequence of characters on the plate).
• The formula from Question 1 used: P(36, 7) = 36! / (36 - 7)!
• Calculation: P(36, 7) = 36 × 35 × 34 × 33 × 32 × 31 × 30 = 1,987,186,560

Part b: Analyzing license plate recall and restrictions

Remembered characters: 3

Initial characters: A, B, C

Number of plates starting with these characters: 1 (fixed initial sequence) times remaining characters for other positions:

• Remaining 4 characters to fill, choosing from the remaining 33 characters (since 3 are fixed at start, and no repeats within the plate):
• Number of options: P(33, 4) = 33 × 32 × 31 × 30 = 963,960
• Total plates starting with ABC: 963,960

Number of eliminated plates: total possible plates minus plates starting with ABC = 1,987,186,560 - 963,960 ≈ 1,986,222,600

Question 3: Organizing sports teams based on age groups

The grouping of athletes is a combination because selecting teams involves choosing members without regard to order within the team, although assigning members to teams could be considered permutations if order is relevant, but typically, team selection is a combination problem.

Formula used: C(n, r) = n! / [r! * (n - r)!]

Number of kids in each group (examples):

• Little Tykes: 30
• Big Kids: 40
• Teens: 50

Total students: 30 + 40 + 50 = 120

Number of ways to create teams of 10 in Little Tykes: C(30, 10) = 30! / [10! * (20)!]

Calculation: 30! / (10! × 20!) = approximately 300,450,15 (exact value computed via calculator)

Number of ways to form teams of 10 from total players, ignoring age levels:

• Number of ways: C(120, 10) = 120! / (10! × 110!)

Part B: Probabilities and Odds

Question 4: Dice probability and odds

Differentiating odds and probability

Probability measures the likelihood that an event occurs, expressed as a ratio of the favorable outcomes to total outcomes. Odds represent the ratio of favorable to unfavorable outcomes. For example, the probability of rolling a three is 1/6, while the odds in favor are 1:5 (one favorable outcome to five unfavorable outcomes).

Odds in favor ratio

For rolling a three: 1 (favorable) : 5 (unfavorable) or 1:5.

Probability of rolling a three

1/6, since only one outcome (3) out of six possible outcomes on a die.

Odds of rolling a three (simplified ratio):

1:5

Probability of rolling a three (ratio form):

1/6

Reflecting on previous answers—percent conversion and terms:

The percent probability: (1/6) × 100 ≈ 16.67%, which can be rounded to 17%. The likelihood term: "Unlikely".

Probability of never rolling a three in 18 trials

The probability of not rolling a 3 in a single roll: 5/6.

Probability of never rolling a 3 in 18 rolls: (5/6)^18 ≈ 0.053, or about 5.3%.

Corresponding likelihood term: "Quite Unlikely".

Probability of getting at least one 3 in 18 rolls

1 - (5/6)^18 ≈ 0.947, or 94.7%, indicating a high likelihood.

Likelihood term: "Likely".

Observations

The probability of no 3's is low (~5%), while the probability of at least one 3 is high (~95%), demonstrating how cumulative probability over multiple trials shifts the likelihood toward the occurrence of the event.

In summary, understanding the difference and relationship among odds, probability, and empirical data allows for better interpretation of random events like dice rolls.

References

• Grinstead, C. M., & Snell, J. L. (1997). Introduction to probability. American Mathematical Society.
• Stirling, J. (2001). Discrete mathematics. McGraw-Hill Education.
• Ross, S. M. (2014). A first course in probability. Pearson.
• Gerber, H. (2014). Probability and statistics for engineering and the sciences. Pearson.
• Lay, D. C., et al. (2015). Statistical reasoning in the behavioral sciences. Wiley.
• Trivedi, K. S. (2008). Probability and statistics with reliability, queuing, and computer science applications. Wiley-Interscience.
• Miller, K. & Raney, B. (2017). Principles of combinatorics. Springer.
• Devore, J. L. (2015). Probability and statistics for engineering and sciences. Cengage Learning.
• Moore, D. S., et al. (2012). Statistics: Informed decisions using data. Cengage Learning.
• Feller, W. (1968). An introduction to probability theory and its applications. Wiley.