🎲Intro to Probability Unit 3 – Counting: Permutations and Combinations

Counting techniques are essential in probability and statistics, focusing on permutations and combinations. These methods help calculate the number of possible outcomes in various scenarios, from simple arrangements to complex selections with specific conditions. Permutations deal with ordered arrangements, while combinations involve unordered selections. Understanding when to use each technique is crucial for solving problems in fields like cryptography, genetics, and game theory. Mastering these concepts provides a foundation for more advanced probability calculations.

Key Concepts and Definitions

  • Permutations involve arranging objects in a specific order where order matters
  • Combinations involve selecting objects from a group where order does not matter
  • Factorial notation n!n! represents the product of all positive integers less than or equal to nn
    • Example: 5!=5×4×3×2×1=1205! = 5 \times 4 \times 3 \times 2 \times 1 = 120
  • Permutations with repetition allow objects to be chosen more than once
  • Combinations with repetition allow objects to be chosen more than once and order does not matter
  • The binomial coefficient (nk)\binom{n}{k} represents the number of ways to choose kk objects from a set of nn objects
  • The multiplication principle states that if there are mm ways to do one thing and nn ways to do another, there are m×nm \times n ways to do both

Fundamental Counting Principle

  • Multiplicative principle states that if there are n1n_1 ways to perform task 1, n2n_2 ways to perform task 2, ..., and nkn_k ways to perform task kk, then there are n1×n2×...×nkn_1 \times n_2 \times ... \times n_k ways to perform all kk tasks together
    • Example: If there are 3 shirt colors and 4 pant colors, there are 3×4=123 \times 4 = 12 possible outfits
  • Additive principle states that if there are n1n_1 ways to perform task 1 and n2n_2 ways to perform task 2, and the tasks are mutually exclusive, then there are n1+n2n_1 + n_2 ways to perform either task
  • The number of ways to arrange nn distinct objects is n!n!
    • Example: The number of ways to arrange the letters A, B, and C is 3!=3×2×1=63! = 3 \times 2 \times 1 = 6
  • The number of ways to arrange nn objects where there are n1n_1 identical objects of type 1, n2n_2 identical objects of type 2, ..., and nkn_k identical objects of type kk is n!n1!×n2!×...×nk!\frac{n!}{n_1! \times n_2! \times ... \times n_k!}
  • The number of ways to select kk objects from a set of nn objects is (nk)=n!k!(nk)!\binom{n}{k} = \frac{n!}{k!(n-k)!}

Permutations Explained

  • A permutation is an ordered arrangement of objects
  • The number of permutations of nn distinct objects is n!n!
  • The number of permutations of nn objects taken rr at a time is denoted as P(n,r)P(n,r) or nPrnPr and is calculated as n!(nr)!\frac{n!}{(n-r)!}
    • Example: The number of ways to select 3 people from a group of 10 to stand in a line is P(10,3)=10!(103)!=10!7!=720P(10,3) = \frac{10!}{(10-3)!} = \frac{10!}{7!} = 720
  • Permutations with repetition, where objects can be chosen more than once, are calculated as nrn^r
    • Example: The number of 4-digit PIN codes using digits 0-9 is 104=10,00010^4 = 10,000
  • Circular permutations, where arrangements in a circle are considered the same if they can be rotated to match, are calculated as (n1)!(n-1)!
  • Permutations with indistinguishable objects are calculated using the formula n!n1!×n2!×...×nk!\frac{n!}{n_1! \times n_2! \times ... \times n_k!}

Combinations Demystified

  • A combination is a selection of objects where order does not matter
  • The number of combinations of nn objects taken rr at a time is denoted as C(n,r)C(n,r), nCrnCr, or (nr)\binom{n}{r} and is calculated as n!r!(nr)!\frac{n!}{r!(n-r)!}
    • Example: The number of ways to select 3 toppings from a list of 10 is C(10,3)=10!3!(103)!=10!3!7!=120C(10,3) = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!} = 120
  • Combinations with repetition, where objects can be chosen more than once and order does not matter, are calculated as (n+r1r)\binom{n+r-1}{r}
    • Example: The number of ways to select 3 scoops of ice cream from 5 flavors, allowing repetition, is (5+313)=(73)=35\binom{5+3-1}{3} = \binom{7}{3} = 35
  • The binomial theorem states that (x+y)n=k=0n(nk)xnkyk(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k} y^k
  • Pascal's triangle is a triangular array of binomial coefficients where each number is the sum of the two numbers above it
  • The hockey-stick identity states that the sum of any r+1r+1 consecutive binomial coefficients in a diagonal of Pascal's triangle equals the next binomial coefficient in the next diagonal

Permutations vs Combinations: When to Use Which

  • Use permutations when order matters and each object can only be used once
    • Example: Arranging books on a shelf, where the order of the books is important
  • Use combinations when order does not matter and each object can only be used once
    • Example: Selecting a committee from a group of people, where the order of selection does not matter
  • Use permutations with repetition when order matters and objects can be chosen more than once
    • Example: Creating a password where characters can be repeated
  • Use combinations with repetition when order does not matter and objects can be chosen more than once
    • Example: Selecting a handful of candy from a bowl, where the order of selection does not matter and candies of the same type are indistinguishable
  • In general, if the problem involves arranging or ranking, use permutations; if the problem involves selecting or grouping, use combinations

Common Pitfalls and How to Avoid Them

  • Not distinguishing between permutations and combinations
    • Ask yourself if the order matters and if objects can be repeated
  • Confusing nn and rr in the formulas
    • Remember that nn represents the total number of objects and rr represents the number of objects being selected or arranged
  • Forgetting to account for repetition or indistinguishable objects
    • Pay attention to whether objects can be chosen more than once or if some objects are identical
  • Overcounting or undercounting possibilities
    • Break the problem down into smaller parts and consider all possible cases
  • Misinterpreting word problems
    • Carefully read the problem and identify the key information and constraints
  • Not simplifying factorials before calculating
    • Cancel out common factors in the numerator and denominator to simplify the calculation
  • Attempting to solve problems using brute force instead of formulas
    • Recognize patterns and use the appropriate formulas to save time and reduce errors

Real-World Applications

  • Lottery and gambling probabilities
    • Example: Calculating the odds of winning a lottery jackpot
  • Cryptography and password security
    • Example: Determining the strength of a password based on the number of possible combinations
  • DNA sequencing and genetic analysis
    • Example: Estimating the number of possible DNA sequences of a given length
  • Resource allocation and scheduling
    • Example: Assigning tasks to employees in different orders to optimize productivity
  • Quality control and sampling
    • Example: Selecting a random sample of products to test for defects
  • Sports and gaming strategies
    • Example: Analyzing the number of possible plays or moves in a game
  • Marketing and product design
    • Example: Creating unique product configurations based on available options
  • Logistics and transportation optimization
    • Example: Determining the most efficient route for a delivery truck to visit multiple locations

Practice Problems and Solutions

  1. How many ways can 5 books be arranged on a shelf?

    • Solution: 5!=5×4×3×2×1=1205! = 5 \times 4 \times 3 \times 2 \times 1 = 120
  2. In how many ways can a committee of 3 people be selected from a group of 10?

    • Solution: C(10,3)=10!3!(103)!=10!3!7!=120C(10,3) = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!} = 120
  3. How many 4-digit PIN codes can be created using digits 0-9 if repetition is allowed?

    • Solution: 104=10,00010^4 = 10,000
  4. How many ways can 5 people be seated around a circular table?

    • Solution: (51)!=4!=24(5-1)! = 4! = 24
  5. A box contains 5 red balls and 3 blue balls. In how many ways can 4 balls be selected if at least 2 must be red?

    • Solution: C(5,2)×C(6,2)+C(5,3)×C(5,1)+C(5,4)=10×15+10×5+5=205C(5,2) \times C(6,2) + C(5,3) \times C(5,1) + C(5,4) = 10 \times 15 + 10 \times 5 + 5 = 205
  6. How many unique 5-character strings can be formed using the letters A, B, and C if repetition is allowed?

    • Solution: 35=2433^5 = 243
  7. In how many ways can 6 identical pens be distributed among 4 people?

    • Solution: (6+4141)=(93)=84\binom{6+4-1}{4-1} = \binom{9}{3} = 84


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© 2024 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.