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1.9 Permutations, Combinations, Probability & Statistics

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Ateneo College Entrance Test ACET

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1 Chapter 1: Mathematics

1.1 Real Numbers, Ratios, Rates & Percentages 1.2 Algebraic Expressions, Polynomials & Linear Equations 1.3 Functions, Domain/Range, Graphs & Transformations 1.4 Quadratic, Polynomial, Exponential & Logarithmic Equations 1.5 Plane Geometry: Triangles, Circles, Polygons & Perimeter/Area 1.6 Coordinate Geometry: Lines, Distance, Midpoints & Conic Sections 1.7 Trigonometry: Right Triangle Ratios, Unit Circle & Basic Identities 1.8 Arithmetic & Geometric Sequences, Series & Word Patterns 1.9 Permutations, Combinations, Probability & Statistics

2 Chapter 2: English Proficiency

2.1 Subject-Verb Agreement, Verb Tenses & Aspect 2.2 Pronoun-Antecedent Agreement, Case, Misplaced & Dangling Modifiers 2.3 Sentence Structure, Run-Ons, Fragments & Parallelism 2.4 Contextual Vocabulary, Word Roots, Prefixes & Suffixes 2.5 Advanced Synonyms, Antonyms & Nuanced Vocabulary 2.6 Verbal Analogies: Relationship Types & Solving Strategies 2.7 Punctuation, Capitalization, Mechanics & Common Errors 2.8 Diction, Idiomatic Expressions, Prepositions & Frequently Confused Words

3 Chapter 3: Reading Comprehension

3.1 Identifying Main Ideas, Central Themes & Paragraph Summaries 3.2 Locating Explicit Details, Supporting Evidence & Data 3.3 Drawing Logical Inferences & Unstated Conclusions 3.4 Analyzing Author's Purpose, Point of View, Tone & Attitude 3.5 Interpreting Figurative Language, Tone Words & Context Clues 3.6 Text Structure, Rhetorical Devices & Evaluating Arguments

4 Chapter 4: Logical and Abstract Reasoning

4.1 Number & Letter Series Patterns & Formulas 4.2 Abstract Reasoning: Matrix Patterns & Figure Rotations 4.3 Formal Syllogisms, Categorical Logic & Deductive Proofs 4.4 Coding-Decoding, Cipher Patterns & Symbol Operations 4.5 Spatial Visualization, Paper Folding & Venn Diagram Logic

5 Chapter 5: General Knowledge

5.1 Philippine History, Government, Constitution & Civics 5.2 Integrated Science: Biology, Chemistry, Physics & Earth Science 5.3 Philippine & World Geography, Basic Economics & Cultural Facts 5.4 Current Events, World Landmarks, Literature & General Trivia

6 Chapter 6: ACET Essay Writing & Exam Strategies

6.1 ACET Essay: Thesis Development, Structuring & Argumentation 6.2 Time Allocation, Pacing Strategies & Subtest Tactics 6.3 Guessing Strategies (No Penalty), Test Anxiety & Exam Day Readiness

1.9 Permutations, Combinations, Probability & Statistics

Key Takeaways Explain This Topic AI

Differentiate permutations (order matters) from combinations (order does not matter).

Calculate circular permutations using (n-1)! and account for identical item repetitions using n! / (n1! n2! ... nk!).

Apply conditional probability rules P(A|B) = P(A and B) / P(B) and addition rules for non-mutually exclusive events.

Utilize standard deviation and z-scores z = (x - mu) / sigma to analyze normally distributed datasets.

Last updated: July 2026

Permutations, Combinations, Probability & Statistics

Combinatorics, probability, and data interpretation represent a critical component of the ACET. Rapid distinction between ordered arrangements and unordered subsets is required.

  • 1. Fundamental Counting Principle & Permutations
  • Fundamental Counting Principle (FCP)

If Event 1 can occur in m m m ways and Event 2 can occur independently in n n n ways, the total number of combined outcomes is m × n m \times n m × n .

Permutations (Order Matters!)

A permutation is an ordered arrangement of elements. Permutations of n items taken k at a time: P ( n , k ) = n ! ( n − k ) ! \textbf{Permutations of } n \textbf{ items taken } k \textbf{ at a time:} \quad P(n, k) = \frac{n!}{(n - k)!} Permutations of n items taken k at a time: P ( n , k ) = ( n − k )! n ! ​ Permutations with Repetition (Distinguishable): P = n ! n 1 ! ⋅ n 2 ! ⋯ n k ! \textbf{Permutations with Repetition (Distinguishable):} \quad P = \frac{n!}{n_1! \cdot n_2! \cdots n_k!} Permutations with Repetition (Distinguishable): P = n 1 ​ ! ⋅ n 2 ​ ! ⋯ n k ​ ! n ! ​ Circular Permutations of n distinct items: P circular = ( n − 1 ) ! \textbf{Circular Permutations of } n \textbf{ distinct items:} \quad P_{\text{circular}} = (n - 1)! Circular Permutations of n distinct items: P circular ​ = ( n − 1 )! Circular Permutations (Keyrings/Necklaces): P keyring = ( n − 1 ) ! 2 \textbf{Circular Permutations (Keyrings/Necklaces):} \quad P_{\text{keyring}} = \frac{(n - 1)!}{2} Circular Permutations (Keyrings/Necklaces): P keyring ​ = 2 ( n − 1 )! ​

2. Combinations (Order Does NOT Matter!)

A combination is a selection of items without regard to order. Combinations of n items taken k at a time: C ( n , k ) = ( n k ) = n ! k ! ( n − k ) ! \textbf{Combinations of } n \textbf{ items taken } k \textbf{ at a time:} \quad C(n, k) = \binom{n}{k} = \frac{n!}{k!(n - k)!} Combinations of n items taken k at a time: C ( n , k ) = ( k n ​ ) = k ! ( n − k )! n ! ​

  • Key Combinatorial Properties
  • ( n k ) = ( n n − k ) \binom{n}{k} = \binom{n}{n - k} ( k n ​ ) = ( n − k n ​ )
  • ( n 0 ) = ( n n ) = 1 \binom{n}{0} = \binom{n}{n} = 1 ( 0 n ​ ) = ( n n ​ ) = 1

Pascal's Identity: ( n k ) + ( n k + 1 ) = ( n + 1 k + 1 ) \binom{n}{k} + \binom{n}{k + 1} = \binom{n + 1}{k + 1} ( k n ​ ) + ( k + 1 n ​ ) = ( k + 1 n + 1 ​ )

3. Probability Axioms & Compound Rules

Theoretical probability of event A A A in sample space S S S : P ( A ) = n ( A ) n ( S ) , 0 ≤ P ( A ) ≤ 1 P(A) = \frac{n(A)}{n(S)}, \quad 0 \le P(A) \le 1 P ( A ) = n ( S ) n ( A ) ​ , 0 ≤ P ( A ) ≤ 1

  • Probability Rules
  • Complement Rule: P ( A ′ ) = 1 − P ( A ) P(A') = 1 - P(A) P ( A ′ ) = 1 − P ( A )

Addition Rule (Any two events): P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A ∩ B ) P(A \cup B) = P(A) + P(B) - P(A \cap B) P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A ∩ B )

Addition Rule (Mutually Exclusive): P ( A ∪ B ) = P ( A ) + P ( B ) P(A \cup B) = P(A) + P(B) P ( A ∪ B ) = P ( A ) + P ( B )

Multiplication Rule (Independent Events): P ( A ∩ B ) = P ( A ) ⋅ P ( B ) P(A \cap B) = P(A) \cdot P(B) P ( A ∩ B ) = P ( A ) ⋅ P ( B )

Conditional Probability: P ( A ∣ B ) = P ( A ∩ B ) P ( B ) ( P ( B ) > 0 ) P(A \mid B) = \frac{P(A \cap B)}{P(B)} \quad (P(B) > 0) P ( A ∣ B ) = P ( B ) P ( A ∩ B ) ​ ( P ( B ) > 0 )

  • 4. Descriptive Statistics & Normal Distribution
  • Measures of Central Tendency
  • Mean ( x ˉ \bar{x} x ˉ ): x ˉ = ∑ x i N \bar{x} = \frac{\sum x_i}{N} x ˉ = N ∑ x i ​ ​

Weighted Mean: x ˉ w = ∑ ( w i x i ) ∑ w i \bar{x}_w = \frac{\sum (w_i x_i)}{\sum w_i} x ˉ w ​ = ∑ w i ​ ∑ ( w i ​ x i ​ ) ​

Median: Middle value when dataset is sorted.

Mode: Most frequently occurring value.

  • Measures of Dispersion & Z-Scores
  • Range = X max − X min = X_{\text{max}} - X_{\text{min}} = X max ​ − X min ​

Standard Deviation ( σ \sigma σ ): σ = ∑ ( x i − μ ) 2 N \sigma = \sqrt{\frac{\sum (x_i - \mu)^2}{N}} σ = N ∑ ( x i ​ − μ ) 2 ​ ​

  • Standardized Z-Score: z = x − μ σ z = \frac{x - \mu}{\sigma} z = σ x − μ ​
  • Empirical Rule (68-95-99.7 Rule for Normal Distributions)
  • ≈ 68 % \approx 68\% ≈ 68% of data lies within μ ± 1 σ \mu \pm 1\sigma μ ± 1 σ
  • ≈ 95 % \approx 95\% ≈ 95% of data lies within μ ± 2 σ \mu \pm 2\sigma μ ± 2 σ
  • ≈ 99.7 % \approx 99.7\% ≈ 99.7% of data lies within μ ± 3 σ \mu \pm 3\sigma μ ± 3 σ
  • Worked Step-by-Step ACET Exam Problems
  • Problem 1: Committee Selection with Group Constraints

A committee of 5 people is to be chosen from a group of 6 men and 4 women. In how many ways can the committee be formed if it must contain at least 3 women?

Step 1: Break down into mutually exclusive cases. Since the committee has 5 members, "at least 3 women" means either 3 women OR 4 women.

Step 2: Case 1 — Exactly 3 women and 2 men.

Ways to select 3 women from 4: ( 4 3 ) = 4 \binom{4}{3} = 4 ( 3 4 ​ ) = 4 .

Ways to select 2 men from 6: ( 6 2 ) = 6 × 5 2 = 15 \binom{6}{2} = \frac{6 \times 5}{2} = 15 ( 2 6 ​ ) = 2 6 × 5 ​ = 15 . Ways for Case 1 = 4 × 15 = 60 \text{Ways for Case 1} = 4 \times 15 = 60 Ways for Case 1 = 4 × 15 = 60

Step 3: Case 2 — Exactly 4 women and 1 man.

Ways to select 4 women from 4: ( 4 4 ) = 1 \binom{4}{4} = 1 ( 4 4 ​ ) = 1 .

Ways to select 1 man from 6: ( 6 1 ) = 6 \binom{6}{1} = 6 ( 1 6 ​ ) = 6 . Ways for Case 2 = 1 × 6 = 6 \text{Ways for Case 2} = 1 \times 6 = 6 Ways for Case 2 = 1 × 6 = 6

Step 4: Sum all mutually exclusive cases. Total Ways = 60 + 6 = 66 \text{Total Ways} = 60 + 6 = 66 Total Ways = 60 + 6 = 66

Problem 2: Conditional Probability Without Replacement

A box contains 5 red marbles and 3 blue marbles. Two marbles are drawn sequentially without replacement. What is the probability that both marbles drawn are red?

Step 1: Determine probability of 1st draw. P ( R 1 ) = 5 5 + 3 = 5 8 P(R_1) = \frac{5}{5 + 3} = \frac{5}{8} P ( R 1 ​ ) = 5 + 3 5 ​ = 8 5 ​

Step 2: Determine conditional probability of 2nd draw given 1st draw was red. Remaining red marbles = 4 = 4 = 4 ; total remaining marbles = 7 = 7 = 7 . P ( R 2 ∣ R 1 ) = 4 7 P(R_2 \mid R_1) = \frac{4}{7} P ( R 2 ​ ∣ R 1 ​ ) = 7 4 ​

Step 3: Apply multiplication rule for dependent events. P ( R 1 ∩ R 2 ) = P ( R 1 ) ⋅ P ( R 2 ∣ R 1 ) = 5 8 × 4 7 = 20 56 = 5 14 P(R_1 \cap R_2) = P(R_1) \cdot P(R_2 \mid R_1) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14} P ( R 1 ​ ∩ R 2 ​ ) = P ( R 1 ​ ) ⋅ P ( R 2 ​ ∣ R 1 ​ ) = 8 5 ​ × 7 4 ​ = 56 20 ​ = 14 5 ​

ACET Speed Tactics & Traps Summary

Problem Type Common Student Trap ACET Speed Tactic

Combinatorics Using permutations when order doesn't matter If roles/positions are identical, use combination ( n k ) \binom{n}{k} ( k n ​ )

Circular Seating Using n ! n! n ! for circular table arrangements Use ( n − 1 ) ! (n-1)! ( n − 1 )! to account for rotational equivalence

Probability Forgetting to update total count when sampling without replacement Reduce both numerator and denominator by 1 for second draw

Mean Alterations Recalculating mean from scratch when adding constant k k k Adding k k k to every entry increases mean by exactly k k k