Study guide
4.5 Counting Principles: Permutations and Combinations
Adamson University College Entrance Test AdUCET
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1 Introduction & Exam Overview
1.1 AdUCET Overview, Format, and Admission Process 1.2 Test-Taking and Time-Management Strategies
2 Chapter 2: Mathematics I — Numbers, Algebra, and Functions
2.1 Number Sense, Operations, and Fraction–Decimal–Percent Conversions 2.2 Algebra, Polynomials, and Linear/Quadratic Equations 2.3 Inequalities, Absolute Value, and Systems of Equations 2.4 Radicals, Rational Expressions, and Scientific Notation 2.5 Coordinate Geometry, Lines, and Functions
3 Chapter 3: Mathematics II — Geometry, Measurement, and Trigonometry
3.1 Plane Geometry: Angles, Triangles, Polygons, and Circles 3.2 Solid Mensuration, Measurement, and Unit Conversion 3.3 Right-Triangle Trigonometry and Applications
4 Chapter 4: Mathematics III — Sequences, Proportion, and Data
4.1 Arithmetic and Geometric Sequences and Series 4.2 Ratio, Proportion, Percentage, Work, and Mixture Problems 4.3 Descriptive Statistics and Data Interpretation 4.4 Probability: Rules, Compound Events, and Conditional Probability 4.5 Counting Principles: Permutations and Combinations
5 Chapter 5: English I — Reading Comprehension and Vocabulary
5.1 Reading Comprehension: Main Idea, Purpose, and Passage Structure 5.2 Inference, Supporting Detail, and Author's Tone 5.3 Word Roots, Prefixes, and Suffixes 5.4 Vocabulary in Context, Synonyms, Antonyms, and Confusables
6 Chapter 6: English II — Grammar, Usage, and Written Expression
6.1 Subject-Verb Agreement and Verb Tenses 6.2 Pronouns, Modifiers, and Parallel Structure 6.3 Sentence Boundaries, Clauses, and Punctuation 6.4 Sentence Correction, Redundancy, and Idiomatic Usage 6.5 Verbal Analogies and Relationship Bridges 6.6 Paragraph Organization, Coherence, and Transitions
7 Chapter 7: General Science I — Biology and Chemistry
7.1 Cell Biology, Transport, and Cellular Energetics 7.2 Genetics, Heredity, and Molecular Biology 7.3 Human Body Systems, Ecology, and Evolution 7.4 Matter, Phase Changes, and Atomic Structure 7.5 Periodic Trends, Chemical Bonding, Reactions, Acids and Bases
8 Chapter 8: General Science II — Physics and Earth Science
8.1 Mechanics: Motion, Newton's Laws, Work, and Energy 8.2 Waves, Sound, Light, Heat, Electricity, and Magnetism 8.3 Earth's Structure, Plate Tectonics, and the Rock Cycle 8.4 The Atmosphere, Weather, the Water Cycle, and Astronomy
9 Chapter 9: Aptitude and Abstract Reasoning
9.1 Figure Series and Pattern Progression 9.2 Matrix Reasoning and Figural Analogies 9.3 Spatial Reasoning: Rotation, Reflection, and Chirality 9.4 Paper Folding, Cube Nets, and Orthographic Views 9.5 Deductive Reasoning, Syllogisms, and Venn Diagrams 9.6 Number, Letter, and Alphanumeric Series 9.7 Odd-One-Out, Classification, and Logic Puzzles
4.5 Counting Principles: Permutations and Combinations
Key Takeaways Explain This Topic AI
Order matters means a permutation, nPr = n!/(n-r)!; order does not matter means a combination, nCr = n!/(r!(n-r)!) = nPr/r!.
The symmetry nCr = nC(n-r) turns hard sums into easy ones: C(10,7) is computed as C(10,3) = 120.
Seating n distinct people around a round table gives (n-1)! arrangements; the same n items on a key ring or necklace give (n-1)!/2 because the ring can be flipped over.
Identical items divide out: MANILA has 6 letters of which two are A, so it has 6!/2! = 360 distinct arrangements.
Count 'must be together' with the block method and 'must not be adjacent' by subtraction: 360 - 120 = 240 MANILA arrangements keep the two A's apart.
Last updated: August 2026
4.5 Counting Principles: Permutations and Combinations
Counting problems on an entrance test are rarely difficult arithmetic. They are difficult classification : almost every mark lost here comes from picking the wrong tool, not from multiplying badly. This section gives you the tools, then the decision rule that selects between them.
1. The Fundamental Counting Principle
The fundamental counting principle (FCP) says that if a task splits into independent stages with m m m , n n n and p p p choices, the number of complete outcomes is m × n × p m \times n \times p m × n × p . A canteen offering 4 kinds of rice, 5 ulam and 3 drinks sells 4 × 5 × 3 = 60 4 \times 5 \times 3 = 60 4 × 5 × 3 = 60 distinct meal combinations.
The FCP also handles restrictions on the fly, one stage at a time. A 4-digit PIN with no repeated digit has 10 × 9 × 8 × 7 = 5,040 10 \times 9 \times 8 \times 7 = 5{,}040 10 × 9 × 8 × 7 = 5 , 040 possibilities, because each stage removes one digit from the pool. If repetition were allowed, the count would be 10 4 = 10,000 10^4 = 10{,}000 1 0 4 = 10 , 000 instead. Whenever repetition is permitted, the answer is a power, never a permutation.
2. Factorials and Linear Permutations
n n n factorial is n ! = n ( n − 1 ) ( n − 2 ) ⋯ ( 2 ) ( 1 ) n! = n(n-1)(n-2)\cdots(2)(1) n ! = n ( n − 1 ) ( n − 2 ) ⋯ ( 2 ) ( 1 ) , with the convention 0 ! = 1 0! = 1 0 ! = 1 . It counts the arrangements of n n n distinct objects in a row. Memorise the small values: 1 , 2 , 6 , 24 , 120 , 720 , 5040 , 40320 1, 2, 6, 24, 120, 720, 5040, 40320 1 , 2 , 6 , 24 , 120 , 720 , 5040 , 40320 for n = 1 n = 1 n = 1 to 8 8 8 .
A permutation is an ordered selection. Choosing and ordering r r r items from n n n distinct items gives n P r = n ! ( n − r ) ! = n ( n − 1 ) ( n − 2 ) ⋯ ⏟ r factors _nP_r = \frac{n!}{(n-r)!} = \underbrace{n(n-1)(n-2)\cdots}_{r \text{ factors}} n P r = ( n − r )! n ! = r factors n ( n − 1 ) ( n − 2 ) ⋯
Never expand the factorials in full — just multiply r r r descending factors. For example 10 P 4 = 10 × 9 × 8 × 7 = 5,040 _{10}P_4 = 10 \times 9 \times 8 \times 7 = 5{,}040 10 P 4 = 10 × 9 × 8 × 7 = 5 , 040 .
Permutations with repeated items
When some of the n n n objects are indistinguishable , divide out the arrangements you cannot tell apart. With n 1 n_1 n 1 copies of one type, n 2 n_2 n 2 of another, and so on: distinct arrangements = n ! n 1 ! n 2 ! ⋯ n k ! \text{distinct arrangements} = \frac{n!}{n_1!\,n_2!\cdots n_k!} distinct arrangements = n 1 ! n 2 ! ⋯ n k ! n !
The word BATANGAS has 8 letters, three of them A, so there are 8 ! / 3 ! = 40,320 / 6 = 6,720 8!/3! = 40{,}320/6 = 6{,}720 8 ! /3 ! = 40 , 320/6 = 6 , 720 distinct arrangements.
Circular and reflective arrangements
Around a circle only relative position matters, because rotating everyone one seat clockwise produces the same arrangement. Fix one person and arrange the rest: circular permutations = ( n − 1 ) ! \text{circular permutations} = (n-1)! circular permutations = ( n − 1 )!
Eight friends around a round table can be seated in 7 ! = 5,040 7! = 5{,}040 7 ! = 5 , 040 ways. If the arrangement can also be flipped over — keys on a key ring, beads on a necklace or bracelet — then each arrangement and its mirror image are the same physical object, so halve the count: reflective circular = ( n − 1 ) ! 2 \text{reflective circular} = \frac{(n-1)!}{2} reflective circular = 2 ( n − 1 )!
Six distinct charms on a bracelet therefore admit 5 ! / 2 = 120 / 2 = 60 5!/2 = 120/2 = 60 5 ! /2 = 120/2 = 60 arrangements.
3. Combinations
A combination is an unordered selection: n C r = n ! r ! ( n − r ) ! = n P r r ! _nC_r = \frac{n!}{r!\,(n-r)!} = \frac{_nP_r}{r!} n C r = r ! ( n − r )! n ! = r ! n P r
The division by r ! r! r ! is the entire difference from a permutation: it erases the r ! r! r ! orderings of each chosen group. Two properties save time on a calculator-free paper:
Symmetry: n C r = n C n − r _nC_r = {}_nC_{n-r} n C r = n C n − r . Choosing 7 of 10 is the same as rejecting 3 of 10, so 10 C 7 = 10 C 3 = 10 × 9 × 8 3 × 2 × 1 = 120 _{10}C_7 = {}_{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 10 C 7 = 10 C 3 = 3 × 2 × 1 10 × 9 × 8 = 120 .
Edges: n C 0 = n C n = 1 _nC_0 = {}_nC_n = 1 n C 0 = n C n = 1 and n C 1 = n _nC_1 = n n C 1 = n .
4. The Decision Table
The question asks for... Does order matter? Tool Formula
Passwords, rankings, podium finishes, officer slates Yes Linear permutation n ! / ( n − r ) ! n!/(n-r)! n ! / ( n − r )!
Committees, teams, card hands, handshakes, chords No Combination n ! / ( r ! ( n − r ) ! ) n!/(r!(n-r)!) n ! / ( r ! ( n − r )!)
Letters of a word with repeated letters Yes Multiset permutation n ! / ( n 1 ! n 2 ! ⋯ ) n!/(n_1!\,n_2!\cdots) n ! / ( n 1 ! n 2 ! ⋯ )
Seating around a round table Relative only Circular permutation ( n − 1 ) ! (n-1)! ( n − 1 )!
Keys on a ring, beads on a necklace Relative, flippable Reflective circular ( n − 1 ) ! / 2 (n-1)!/2 ( n − 1 )! /2
Independent stages: menu, PIN, plate number Yes, with repeats allowed FCP m × n × p m \times n \times p m × n × p
The fastest test: swap two of the chosen items. If you now have a genuinely different result, order matters (permutation); if it is the same result, it does not (combination). President-and-Secretary Ana-then-Ben differs from Ben-then-Ana, so that is a permutation. A committee of Ana and Ben is the same committee either way, so that is a combination.
5. Worked Example 1 — a committee with restrictions
Problem. A student council has 12 eligible members: 7 women (including Ana) and 5 men (including Ben). A 5-member committee is to be formed.
(a) How many committees are possible? Order is irrelevant: 12 C 5 = 12 × 11 × 10 × 9 × 8 5 × 4 × 3 × 2 × 1 = 95,040 120 = 792 _{12}C_5 = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = \frac{95{,}040}{120} = 792 12 C 5 = 5 × 4 × 3 × 2 × 1 12 × 11 × 10 × 9 × 8 = 120 95 , 040 = 792
(b) Exactly 3 women and 2 men? Choose within each pool, then multiply by the FCP: 7 C 3 × 5 C 2 = 35 × 10 = 350 _7C_3 \times {}_5C_2 = 35 \times 10 = 350 7 C 3 × 5 C 2 = 35 × 10 = 350
(c) At least 4 women? Split into cases and add: ( 7 C 4 × 5 C 1 ) + ( 7 C 5 × 5 C 0 ) = ( 35 × 5 ) + ( 21 × 1 ) = 175 + 21 = 196 ({}_7C_4 \times {}_5C_1) + ({}_7C_5 \times {}_5C_0) = (35 \times 5) + (21 \times 1) = 175 + 21 = 196 ( 7 C 4 × 5 C 1 ) + ( 7 C 5 × 5 C 0 ) = ( 35 × 5 ) + ( 21 × 1 ) = 175 + 21 = 196
(d) The committee must include Ana. Ana's seat is fixed, so choose the other 4 from the remaining 11 people: 11 C 4 = 11 × 10 × 9 × 8 24 = 7,920 24 = 330 _{11}C_4 = \frac{11 \times 10 \times 9 \times 8}{24} = \frac{7{,}920}{24} = 330 11 C 4 = 24 11 × 10 × 9 × 8 = 24 7 , 920 = 330
(e) Ana and Ben cannot both serve. Count the forbidden committees and subtract. If both are in, the other 3 come from the remaining 10 people: 10 C 3 = 120 _{10}C_3 = 120 10 C 3 = 120 . So the allowed total is 792 − 120 = 672 792 - 120 = 672 792 − 120 = 672
Restriction problems almost always yield to one of three moves: fix the required member, split into cases , or subtract the forbidden count from the total . The subtraction route is usually the shortest when the words "not", "at least" or "cannot both" appear.
6. Worked Example 2 — arrangements of a word
Problem. Using all the letters of MANILA , find (a) the number of distinct arrangements, (b) the number in which the two A's are together, and (c) the number in which the two A's are not together.
(a) MANILA has 6 letters: M, A, N, I, L, A. Two of them are A; all the others are unique. 6 ! 2 ! = 720 2 = 360 \frac{6!}{2!} = \frac{720}{2} = 360 2 ! 6 ! = 2 720 = 360
(b) Block method. Glue the two A's into a single unit "AA". You now arrange 5 objects — AA, M, N, I, L — in a row: 5 ! = 120 5! = 120 5 ! = 120
There is no extra internal factor here because the two A's are identical; had the block held two different letters, you would multiply by 2 ! 2! 2 ! for their internal order.
(c) By subtraction. 360 − 120 = 240 360 - 120 = 240 360 − 120 = 240
Gap-method check. Arrange the four non-A letters first: 4 ! = 24 4! = 24 4 ! = 24 ways. That creates 5 gaps (one before each letter, one after the last): _ M _ N _ I _ L _ <- 5 gaps, choose 2 for the A's
Choose 2 of the 5 gaps for the A's, order irrelevant since the A's are identical: 5 C 2 = 10 _5C_2 = 10 5 C 2 = 10 . Total 24 × 10 = 240 24 \times 10 = 240 24 × 10 = 240 , matching the subtraction. Two independent methods agreeing is the best confirmation available on a timed paper.
7. Common Traps
Using a permutation for a committee. Handshakes among 10 people are 10 C 2 = 45 _{10}C_2 = 45 10 C 2 = 45 , not 10 P 2 = 90 _{10}P_2 = 90 10 P 2 = 90 ; the permutation counts each handshake twice.
Forgetting to halve for a key ring. Seven keys on a ring give 6 ! / 2 = 360 6!/2 = 360 6 ! /2 = 360 , not 6 ! = 720 6! = 720 6 ! = 720 .
Losing the internal order of a block. If two distinct people must sit together, multiply by 2 after treating them as one unit.
Forgetting the circular reduction inside a block problem. Six friends at a round table with two who must sit together gives ( 5 − 1 ) ! × 2 = 48 (5-1)! \times 2 = 48 ( 5 − 1 )! × 2 = 48 , not 5 ! × 2 = 240 5! \times 2 = 240 5 ! × 2 = 240 .
Attacking "not adjacent" head-on. Count the total, count the together-cases, subtract.
Ignoring repeated letters. Dividing by the factorials of the repeat counts is not optional.
Using n P r _nP_r n P r when repetition is allowed. Repetition means n r n^r n r ; the descending-factor formula assumes each item is used once.
Adding cases that overlap. When splitting "at least 4 women" into cases, the cases must be mutually exclusive by construction — 4 women and 5 women cannot both describe the same committee.