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Counting Principles

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Counting Principles

🎯 Learning Objectives

  • Apply the product rule and sum rule for counting
  • Compute permutations (P(n,k)P(n,k)) and combinations (C(n,k)C(n,k))
  • Understand binomial coefficients and Pascal's identity
  • Distinguish permutations with vs. without repetition
  • Solve counting problems with multiple cases

1.1 Intuition: Counting Without Listing

Counting is the oldest mathematical art. The key insight: we can count things without listing them by understanding the structure of choices.
🔑 Key Insight: Most counting problems reduce to: "How many ways to make a sequence of choices?"

1.2 Basic Rules

Product Rule

If task AA can be done in mm ways and task BB in nn ways, then AA AND BB can be done in m×nm \times n ways. Example: How many 2-letter words? 26×26=67626 \times 26 = 676.

Sum Rule

If task AA can be done in mm ways and task BB in nn ways, and they are disjoint, then AA OR BB can be done in m+nm + n ways. Example: Choosing a vowel (55) or a consonant (2121) from alphabet: 5+21=265 + 21 = 26.

1.3 Permutations

Permutation: An ordered arrangement of distinct objects.
P(n,k)=n×(n1)××(nk+1)=n!(nk)!P(n, k) = n \times (n-1) \times \cdots \times (n-k+1) = \frac{n!}{(n-k)!}
Example: Number of ways to award gold, silver, bronze to 10 athletes: P(10,3)=10×9×8=720P(10,3) = 10 \times 9 \times 8 = 720.

Permutations with Repetition

nkn^k — choose from nn options, kk times, with repetition allowed. Example: Number of 4-digit PINs: 104=1000010^4 = 10000.

1.4 Combinations

Combination: An unordered selection of distinct objects.
C(n,k)=(nk)=n!k!(nk)!C(n, k) = \binom{n}{k} = \frac{n!}{k!(n-k)!}
Example: Number of ways to choose a 3-person committee from 10: (103)=120\binom{10}{3} = 120.

Relationship

P(n,k)=C(n,k)×k!P(n,k) = C(n,k) \times k!

1.5 Binomial Coefficients

(x+y)n=k=0n(nk)xnkyk(x + y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k} y^k

Pascal's Identity

(nk)=(n1k1)+(n1k)\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}

📊 Formula Summary

ConceptFormula
PermutationsP(n,k)=n!(nk)!P(n,k) = \frac{n!}{(n-k)!}
CombinationsC(n,k)=(nk)=n!k!(nk)!C(n,k) = \binom{n}{k} = \frac{n!}{k!(n-k)!}
Permutations with repetitionnkn^k
Binomial theorem(x+y)n=(nk)xnkyk(x+y)^n = \sum \binom{n}{k} x^{n-k} y^k

✅ Practice Questions

Q1: How many ways to arrange the letters in "MATH"?
Solution
4 distinct letters: 4!=244! = 24 arrangements. Q2: How many 5-card poker hands from a 52-card deck? Solution
(525)=52!5!47!=2,598,960\binom{52}{5} = \frac{52!}{5!47!} = 2,598,960. Q3: How many ways to choose a president, VP, and secretary from 12 people? Solution
P(12,3)=12×11×10=1320P(12,3) = 12 \times 11 \times 10 = 1320. Q4: Prove C(n,k)=C(n,nk)C(n,k) = C(n, n-k) combinatorially. Solution
Choosing kk elements to include is equivalent to choosing nkn-k elements to exclude. So (nk)=(nnk)\binom{n}{k} = \binom{n}{n-k}. Q5: How many bit strings of length 8 contain exactly three 1s? Solution
Choose 3 positions out of 8 for the 1s: (83)=56\binom{8}{3} = 56. Join Discord NextAdvanced Counting
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