What is the difference between permutations and combinations?

The difference is whether order matters. In a permutation, the arrangement of items is significant — selecting A, B, C is different from C, B, A. The formula is P(n,r) = n! / (n-r)!. In a combination, the group of items is all that matters — selecting A, B, C is identical to C, B, A. The formula is C(n,r) = n! / (r!(n-r)!). A quick rule: if you can imagine the same group being 'rearranged' into a different valid outcome, you need a permutation. If all rearrangements are equivalent, use a combination.

What is the fundamental counting principle?

The Fundamental Counting Principle (also called the Multiplication Rule of Counting) states that if one event can happen in m ways and a second independent event can happen in n ways, then both events together can happen in m × n ways. For example, if you have 3 shirt choices and 4 pants choices, you can create 3 × 4 = 12 different outfits.

What is a factorial and why is it used in counting?

A factorial (written n!) is the product of all positive integers from 1 up to n. For example, 5! = 5 × 4 × 3 × 2 × 1 = 120. Factorials appear in permutation and combination formulas because they count all possible arrangements. An important special case is 0! = 1, which is defined by mathematical convention and is essential for formulas involving choosing zero items from a group.

When should I use permutations vs combinations?

Use permutations when the order of selection matters: ranking contestants, assigning positions (President, VP, Secretary), forming codes where digit order determines the result. Use combinations when only the group matters: choosing a committee, selecting lottery numbers, dealing a hand of cards. The simplest test: ask whether swapping two of the selected items would create a 'different' outcome. If yes, use permutations. If no, use combinations.

How many arrangements does the word MISSISSIPPI have?

MISSISSIPPI has 11 letters with repeated letters: 1 M, 4 I's, 4 S's, and 2 P's. The formula for permutations with repeated items is n! / (n1! × n2! × ... × nk!). So the number of distinct arrangements is 11! / (1! × 4! × 4! × 2!) = 39,916,800 / (1 × 24 × 24 × 2) = 34,650 unique arrangements.

What is the counting principle in probability?

In probability, counting methods determine the size of the sample space (total outcomes) and the size of an event (favorable outcomes). The probability of an event is then P(event) = favorable outcomes / total outcomes. For example, counting methods tell you there are C(52,5) = 2,598,960 possible 5-card poker hands, which becomes the denominator when calculating the probability of any specific hand type.

Is a combination lock actually a combination or a permutation?

A standard combination lock is actually a permutation lock. The reason: order matters. If the code is 12-34-56, entering 34-12-56 will not open the lock. Because different sequences of the same numbers produce different results, this is a permutation problem, not a combination problem. The lock is traditionally misnamed.

How do you use the Fundamental Counting Principle for a license plate?

A typical license plate with 3 letters followed by 3 digits (repetition allowed) has 26 choices for each letter position and 10 choices for each digit position. Applying the Fundamental Counting Principle: 26 × 26 × 26 × 10 × 10 × 10 = 17,576,000 possible plates. If repetition were not allowed, the calculation would be 26 × 25 × 24 × 10 × 9 × 8 = 11,232,000 possible plates.

Counting Methods: Formulas, Rules & Worked Examples (2026)

Q: What are counting methods in math?

Counting methods in math are techniques for determining the total number of possible outcomes or arrangements without listing every single one. The three core types are: (1) The Fundamental Counting Principle — multiply the number of choices at each stage; (2) Permutations — count ordered arrangements where order matters; (3) Combinations — count selections where order does not matter.

Q: What is n choose r (nCr)?

nCr — read as 'n choose r' — is the combination formula that calculates how many ways you can select r items from a total pool of n items when order does not matter. The formula is C(n,r) = n! / (r! × (n-r)!). For example, C(10,3) = 10! / (3! × 7!) = 120 tells you there are 120 ways to choose a 3-person committee from 10 people.

What Are Counting Methods? (Definition + Overview)

Definition — Counting Methods

Counting methods are mathematical techniques for determining the number of possible outcomes, arrangements, or selections from a set of items — without writing out every possibility individually. They sit at the foundation of probability, discrete mathematics, and computer science, because you cannot calculate P(event) = favorable / total without first knowing how large your total sample space is.

Total outcomes = product of choices at each stage

The field is formally studied under combinatorics — a branch of mathematics concerned with counting, arranging, and selecting objects. According to the treatment of combinatorics in OpenStax Discrete Mathematics, counting principles provide the essential bridge between listing outcomes one by one and computing them systematically. The same ideas appear in MIT's OpenCourseWare treatment of discrete mathematics (6.042J), where counting and combinatorics form a dedicated unit because they are prerequisite to probability, cryptography, and algorithm analysis.

⚡ Quick Reference — The Four Core Counting Methods

Fundamental Counting Principle: m × n × p × … — multiply the number of options at each independent stage
Permutations (no repetition): P(n,r) = n! / (n−r)! — ordered selections where each item can only be chosen once
Permutations (with repetition): n^r — ordered selections where items may repeat
Combinations (no repetition): C(n,r) = n! / (r!(n−r)!) — unordered selections where each item can only be chosen once
Key question: Does changing the arrangement create a different outcome? Yes → permutation. No → combination.
0! = 1 by definition — this is not optional, it makes the formulas work for r = n and r = 0.

Core counting methods

Factorial — the engine of all formulas

Questions that choose the right formula

0! = 1

Critical edge case — never forget

Factorials: The Engine Behind Every Formula

Before covering the four methods, you need to understand the factorial, because it appears in every permutation and combination formula.

Definition — Factorial

n! = n × (n−1) × (n−2) × … × 2 × 1

Read as "n factorial." It is the product of every positive integer from 1 up to n. It counts the number of ways to arrange n distinct objects in a line.

n = total number of items 5! = 120 0! = 1 (by definition)

Here are the factorials you will encounter most often. Memorizing up to 7! saves time on standardized tests:

n	n! (expanded)	Value	Common use
0!	defined as 1	1	Makes formulas work when r = n or r = 0
1!	1	1	Trivial case
2!	2 × 1	2	Arrangements of 2 items
3!	3 × 2 × 1	6	Arrangements of 3 items
4!	4 × 3 × 2 × 1	24	Card suits, small groups
5!	5 × 4 × 3 × 2 × 1	120	Permutations of 5 items
6!	6 × 5 × 4 × 3 × 2 × 1	720	Die faces, seating 6 people
7!	7 × 6 × 5 × 4 × 3 × 2 × 1	5,040	Common in combination problems
10!	10 × 9 × … × 1	3,628,800	Arranging 10 distinct objects

⚠️

Stop! The 0! = 1 Rule Is Not Optional

Students routinely write 0! = 0 and break every formula that uses it. By mathematical convention, 0! = 1. The intuitive reason: there is exactly one way to arrange zero objects — do nothing. Every permutation and combination formula relies on this. When r = n in P(n,r), the denominator becomes (n−n)! = 0! = 1, so P(n,n) = n! / 1 = n!. This is correct: there are n! ways to arrange all n items.

The Fundamental Counting Principle

Definition and Formula

Fundamental Counting Principle (Multiplication Rule)

Total outcomes = n₁ × n₂ × n₃ × … × nₖ

If a task has k sequential stages, and stage 1 can happen in n₁ ways, stage 2 in n₂ ways, and so on independently, then the total number of ways to complete the entire task is the product of all stage counts.

Each stage must be independent Order of stages matters Repetition is allowed by default

This is the most fundamental rule in all of counting, and the others build on top of it. The key word is independent: the number of choices at stage 2 must not change depending on what happened at stage 1 (or you need to adjust the count at each stage).

Worked Example 1 — Outfit Combinations

Fundamental Counting Principle

You have 3 pairs of pants, 4 shirts, and 2 pairs of shoes. How many distinct outfits can you make?

Count each stage independently. Pants: 3 options. Shirts: 4 options. Shoes: 2 options. Each choice does not restrict the others — choosing Shirt A does not eliminate any pants or shoes.

Apply the multiplication rule. Total = 3 × 4 × 2 = 24 distinct outfits.

Verify the logic. For each of the 3 pants choices, there are 4 shirt choices → 12 combinations. For each of those 12 combinations, there are 2 shoe choices → 24 total. A tree diagram would show exactly 24 branches.

✓ Answer: 3 × 4 × 2 = 24 distinct outfits

Worked Example 2 — License Plates

Fundamental Counting Principle — Repetition Allowed

A license plate has 3 letters followed by 3 digits. Repetition is allowed. How many unique plates are possible?

Identify the stages. 3 letter positions (26 choices each) + 3 digit positions (10 choices each). Repetition is allowed, so every position keeps its full range of choices.

Multiply. Total = 26 × 26 × 26 × 10 × 10 × 10 = 26³ × 10³ = 17,576 × 1,000 = 17,576,000.

Compare to no-repetition. If no letter or digit could repeat: 26 × 25 × 24 × 10 × 9 × 8 = 11,232,000. Repetition adds about 6.3 million additional possibilities.

✓ Answer (repetition allowed): 17,576,000 unique license plates

Permutations: When Order Matters

A permutation counts the number of ways to select and arrange r items from a group of n, where the order of selection creates a distinct outcome. Swapping any two chosen items produces a different permutation.

Permutations Without Repetition (the Standard Case)

Permutations — No Repetition

P(n, r) = n! / (n − r)!

Read as "the permutation of n things taken r at a time." It counts ordered arrangements of r items chosen from n distinct items, where no item appears twice. Also written as nPr or ₙPᵣ.

n = total items in the pool r = items being selected and arranged Condition: r ≤ n

Worked Example 3 — Olympic Medal Ceremony

Permutation — No Repetition

Eight runners compete in a sprint final. Gold, Silver, and Bronze medals are awarded. How many different medal-stand outcomes are possible?

Identify n and r. Total runners: n = 8. Medals awarded (positions selected): r = 3. Each runner can only win one medal, so no repetition.

Does order matter? Yes. Runner A winning Gold and Runner B winning Silver is a different outcome from Runner B winning Gold and Runner A winning Silver. This confirms we need a permutation.

Apply the formula. P(8, 3) = 8! / (8−3)! = 8! / 5! = (8 × 7 × 6 × 5!) / 5! = 8 × 7 × 6 = 336.

Intuition check. 8 choices for Gold. For each Gold winner, 7 remaining choices for Silver. For each Silver winner, 6 remaining for Bronze. Fundamental Counting Principle confirms: 8 × 7 × 6 = 336. The formulas agree.

✓ Answer: P(8, 3) = 336 distinct medal-stand outcomes

Permutations With Repetition

Permutations — Repetition Allowed

Total = n^r

When you select r items from n types and items can be reused, there are n choices at every position — so the total is n multiplied by itself r times. This is directly the Fundamental Counting Principle applied to identical stage sizes.

n = number of distinct item types r = number of positions/selections Example: 4-digit PIN = 10⁴ = 10,000

💡

PIN Codes, Lock Codes, and License Plates Use n^r

Any time every position independently draws from the same full pool — a 4-digit PIN from digits 0–9 (10⁴ = 10,000), a 6-character password from 26 lowercase letters (26⁶ = 308,915,776) — you are counting permutations with repetition. The license plate example in Section 3 is also n^r applied per letter group and per digit group.

Permutations With Identical Items

A third permutation scenario arises when you want to arrange all n items but some of them are identical. The standard n! overcounts because swapping two identical letters produces the same word. The correction formula divides out those duplicates:

Permutations — With Identical Items

Total = n! / (n₁! × n₂! × … × nₖ!)

Where n is the total number of items, and n₁, n₂, … nₖ are the counts of each repeated item type. This formula is sometimes called the "multinomial coefficient."

Worked Example 4 — Arranging "MISSISSIPPI"

Permutation — Identical Items

How many distinct arrangements exist for the letters in the word MISSISSIPPI?

Count total letters and each repeated letter. MISSISSIPPI has 11 letters total: 1 M, 4 I's, 4 S's, 2 P's.

Apply the formula. Total = 11! / (1! × 4! × 4! × 2!)

Calculate. 11! = 39,916,800. Denominator = 1 × 24 × 24 × 2 = 1,152. Total = 39,916,800 / 1,152 = 34,650.

Why divide? If all 11 letters were distinct, there would be 11! arrangements. But swapping any two I's, any two S's, or the two P's produces the same word — so those swaps get divided out.

✓ Answer: 11! / (1! × 4! × 4! × 2!) = 34,650 distinct arrangements

This example is a standard problem in combinatorics textbooks. The calculation matches the treatment in Grinstead & Snell's "Introduction to Probability" (Dartmouth), Chapter 3.

Combinations: When Order Does Not Matter

A combination counts the number of ways to select r items from a group of n where the group composition is all that matters. Rearranging the chosen items produces the same combination.

Combinations — No Repetition (nCr)

C(n, r) = n! / (r! × (n − r)!)

Read as "n choose r." It counts how many ways you can select r items from n distinct items when order does not matter. Also written as ₙCᵣ or the binomial coefficient (n choose r) in parentheses.

n = total items in the pool r = items being selected C(n,r) = P(n,r) / r!

The relationship C(n,r) = P(n,r) / r! is worth understanding intuitively: you first count all ordered arrangements (permutations), then divide by r! because every unique group of r items generates r! different orderings — all of which represent the same combination.

✅

Symmetry Property: C(n,r) = C(n, n−r)

Choosing 3 items from 10 to include in a group gives the same count as choosing 7 items to leave out. C(10,3) = C(10,7) = 120. This symmetry property is a useful sanity check and sometimes shortens calculations when r is large.

Worked Example 5 — Choosing a Committee

Combination — No Repetition

A club of 10 people wants to elect a 3-person committee. How many distinct committees are possible?

Does order matter? No. A committee of {Alice, Bob, Carol} is the same committee regardless of who was named first. The roles are identical — there is no "first committee member" versus "second." This confirms a combination.

Identify n and r. n = 10 (total people), r = 3 (committee size).

Apply the formula. C(10, 3) = 10! / (3! × 7!) = (10 × 9 × 8) / (3 × 2 × 1) = 720 / 6 = 120.

✓ Answer: C(10, 3) = 120 distinct 3-person committees

Worked Example 6 — 5-Card Poker Hand

Combination — No Repetition

A standard 52-card deck is shuffled. How many distinct 5-card poker hands are possible?

Does order matter? No. A hand of {A♠, K♥, Q♦, J♣, 10♠} is the same hand no matter what order the cards are dealt. Order of dealing does not change what hand you hold.

Identify n and r. n = 52 (total cards), r = 5 (hand size).

Apply the formula. C(52, 5) = 52! / (5! × 47!) = (52 × 51 × 50 × 49 × 48) / (5 × 4 × 3 × 2 × 1) = 311,875,200 / 120 = 2,598,960.

Connection to probability. This 2,598,960 becomes the denominator when you calculate the probability of any particular hand type. For example, there are 4 Royal Flushes (one per suit), so P(Royal Flush) = 4 / 2,598,960 ≈ 0.000154%.

✓ Answer: C(52, 5) = 2,598,960 distinct 5-card poker hands

Master Formula Cheat Sheet

This table is designed for fast reference during an exam. For each scenario, identify whether order matters and whether repetition is allowed — then look up the formula.

Method	Formula	Order Matters?	Repetition Allowed?	Classic Example
Fundamental Counting Principle	n₁ × n₂ × … × nₖ	Yes (stages are sequential)	Yes (default)	Outfit combinations: 3 × 4 × 2 = 24
Permutation (no repetition)	P(n,r) = n! / (n−r)!	Yes	No	Medal stand: P(8,3) = 336
Permutation (repetition allowed)	n^r	Yes	Yes	4-digit PIN: 10⁴ = 10,000
Permutation (identical items)	n! / (n₁! × n₂! × … × nₖ!)	Yes	Forced by repeated items	MISSISSIPPI: 11!/(1!4!4!2!) = 34,650
Combination (no repetition)	C(n,r) = n! / (r!(n−r)!)	No	No	Committee of 3 from 10: C(10,3) = 120

Interactive Decision Tree: Pick the Right Formula

Two questions are all you need to identify the correct counting method. Work through both before writing down any formula.

📐 Order vs. Repetition — Formula Finder

Does the ORDER of items matter?
Would swapping two items create a different valid outcome?

✅ YES — Order matters

Can items be REPEATED?
Can the same item be chosen more than once?

✅ YES

Permutation
(repetition allowed)
n^r

❌ NO

Permutation
(no repetition)
P(n,r) = n!/(n−r)!

❌ NO — Order does not matter

Can items be REPEATED?
Can the same item be chosen more than once?

✅ YES

Combination
(repetition — stars & bars)
C(n+r−1, r)

❌ NO

Combination
(no repetition)
C(n,r) = n!/(r!(n−r)!)

⚠️

Stop! Double-Check Before Applying a Formula

The single most common error in counting problems is choosing the wrong formula. Always ask both questions explicitly: (1) "Does swapping two of my chosen items produce a different valid outcome?" and (2) "Can the same item be chosen more than once?" Write out your answers before picking a formula. Students who write down the two questions correctly almost always pick the right formula.

The Lock Code Paradox: Why a "Combination Lock" Is a Lie

🔐 Mind-Shift — The Classic Misconception

A standard combination lock is not a combination lock. It is a permutation lock.

Consider a standard padlock with the code 15-30-45. If you enter 45-30-15, the lock stays shut. The sequence matters — a different order of the same three numbers produces a different outcome. By definition, that is a permutation, not a combination.

In a true "combination" problem, the order would not matter. Choosing team members {Alice, Bob, Carol} is a combination — you end up with the same three-person team no matter what order you name them. Opening a lock does not work that way.

How to use this to remember the rule forever: Every time you see the word "combination" in everyday life, ask yourself whether the order matters. If yes, it is really a permutation. If no, it is a combination. The lock is the clearest possible example of how common language conflicts with mathematical precision — and that contrast makes the distinction unforgettable.

Permutations vs. Combinations: Side-by-Side Comparison

Feature	Permutation (P)	Combination (C)
Does order matter?	Yes — different orders = different outcomes	No — all orders = same outcome
Formula (no repetition)	P(n,r) = n! / (n−r)!	C(n,r) = n! / (r!(n−r)!)
Relationship	Always ≥ C(n,r) for same n, r	C(n,r) = P(n,r) / r!
Classic example	Race podium, password, anagram	Committee, card hand, lottery
Keyword signal	"Arrange," "rank," "ordered," "sequence," "assign"	"Choose," "select," "group," "subset," "committee"
P(4,3) vs C(4,3)	P(4,3) = 24	C(4,3) = 4
Symmetry property	No general symmetry	C(n,r) = C(n, n−r)
Counts for C(52,5)	P(52,5) = 311,875,200	C(52,5) = 2,598,960

Tree Diagrams: Visualizing Sequential Choices

A tree diagram draws out every possible outcome of a sequential process as a branching path. They work best for small problems and are a powerful tool for building intuition before you apply a formula.

Tree diagrams confirm the Fundamental Counting Principle visually: the number of final branches (leaves) always equals the product of branches at each stage. For 2 choices at stage 1 and 2 choices at stage 2, you get 2 × 2 = 4 leaf nodes.

Interactive Counting Methods Calculator

🧮 Counting Methods Calculator — Factorial, nPr, nCr

n (integer ≥ 0)

n (total items)

r (items selected)

n (total items)

r (items selected)

Enter the number of choices at each stage (up to 6 stages), separated by commas or one per field.

Stage 1 choices

Stage 2 choices

Stage 3 choices

Stage 4 (optional)

Stage 5 (optional)

Stage 6 (optional)

Counting Methods and Probability

Counting methods and basic probability are inseparable. The probability formula P(A) = favorable outcomes / total outcomes requires you to count both the numerator and denominator before dividing. Without counting methods, you can only calculate probabilities for simple coin-flip experiments. With them, you can handle card games, lottery odds, and combinatorial analysis of complex systems.

AI Overview Block — Counting Methods in Probability

How Counting Methods Map to Probability Calculations

Counting methods determine the size of the sample space (total outcomes) and the size of an event (favorable outcomes). The Fundamental Counting Principle, permutations, and combinations each define different kinds of sample spaces. Once those sizes are known, the probability rules (addition, multiplication, complement) can be applied to any event within that space. For example: C(52,5) = 2,598,960 total poker hands; C(4,1) × C(48,4) = 778,320 hands containing exactly one ace; P(one ace) = 778,320 / 2,598,960 ≈ 29.9%.

This connection between counting and probability is documented throughout the probability literature. The treatment in MIT's 6.042J Mathematics for Computer Science explicitly presents counting methods as prerequisite to probability, noting that computing the probability of any event in a finite sample space reduces to a counting problem. Similarly, the Khan Academy combinatorics unit introduces permutations and combinations specifically as tools for computing probabilities in card, marble, and selection problems.

For further reading on how counting connects to downstream topics covered on this site, see conditional probability, expected value, and the binomial distribution — which is built directly on combination coefficients C(n,k).

Common Mistakes and How to Avoid Them

Mistake	❌ Wrong Approach	✅ Correct Approach
Using 0! = 0	P(5,5) = 5! / 0! = 120 / 0 = undefined	0! = 1 by definition → P(5,5) = 120 / 1 = 120 ✓
Using permutation when order doesn't matter	Committee of 3 from 10: P(10,3) = 720	Committee: C(10,3) = 120 (divides out 3! = 6 orderings)
Using combination when order matters	Top-3 race finish: C(8,3) = 56 outcomes	Race podium: P(8,3) = 336 (different orderings = different results)
Forgetting to adjust for identical items	Arrangements of MISSISSIPPI: 11! = 39,916,800	11! / (1!×4!×4!×2!) = 34,650 (divide out repeated-letter permutations)
Treating independent stages as dependent	2nd digit of PIN affected by 1st digit	Each PIN digit is independent → 10⁴ = 10,000 total (repetition allowed)
Confusing "combination lock" with math	Lock with code 15-30-45 uses combinations	Order matters → it is a permutation lock

Entity and Formula Glossary

This glossary uses standardized notation for AI extraction, LLM readability, and Featured Snippet optimization. Every term is paired with its formula and a plain-text definition.

Term	Notation / Formula	Plain-English Definition
Factorial	n! = n × (n−1) × … × 1	The product of all positive integers from 1 to n. Counts total arrangements of n distinct objects. Special case: 0! = 1.
Fundamental Counting Principle	n₁ × n₂ × … × nₖ	If a task has k independent stages with n₁, n₂, … nₖ choices respectively, total outcomes = their product.
Permutation (no repetition)	P(n,r) = n! / (n−r)!	Ordered selections of r items from n distinct items, each used at most once. Also written nPr.
Permutation (repetition)	n^r	Ordered selections of r items from n types when items may be reused. Each position has n independent choices.
Permutation (identical items)	n! / (n₁! × n₂! × … × nₖ!)	Arrangements of n items where n₁ items are of type 1, n₂ of type 2, etc. Divides out duplicate arrangements.
Combination (nCr)	C(n,r) = n! / (r!(n−r)!)	Unordered selections of r items from n distinct items. Read "n choose r." Equal to C(n, n−r) by symmetry.
Sample Space	S or Ω	The complete set of all possible outcomes in a probability experiment. Counting methods determine \|S\|, its size.
Distinct vs. Indistinguishable	Distinct: all items unique. Indistinguishable: some items identical.	Distinct items → standard P or C formulas. Indistinguishable items → divide by the factorial of each repeated group.
Binomial Coefficient	C(n,r) — same as combination	Called "binomial coefficient" because it appears as the coefficient of x^r in the expansion of (1+x)ⁿ. Foundational to the binomial distribution.

Frequently Asked Questions

Counting methods in math are techniques for finding the total number of possible outcomes or arrangements from a set without listing every single one. The three core types are: (1) the Fundamental Counting Principle — multiply choices at each stage; (2) Permutations — ordered arrangements where order creates a different result; and (3) Combinations — unordered selections where only the group composition matters. They are essential for computing probability, because probability = favorable outcomes / total outcomes.

The difference is whether order matters. Permutation: order matters — {A, B, C} and {C, B, A} are different arrangements. Formula: P(n,r) = n! / (n−r)!. Combination: order does not matter — {A, B, C} and {C, B, A} are the same group. Formula: C(n,r) = n! / (r!(n−r)!). A quick test: if swapping two chosen items creates a different valid outcome, use a permutation. If not, use a combination. Since combinations ignore order, C(n,r) = P(n,r) / r! — combinations are always fewer.

The Fundamental Counting Principle states: if one event can occur in m ways and an independent second event can occur in n ways, then both together can occur in m × n ways. This extends to any number of stages: multiply all the stage counts together. Example: 3 pants choices × 4 shirt choices × 2 shoe choices = 3 × 4 × 2 = 24 distinct outfits. The principle applies whenever stages are independent — the number of choices at one stage does not depend on what was chosen at another.

A factorial (n!) is the product of every positive integer from 1 up to n. For example, 5! = 5 × 4 × 3 × 2 × 1 = 120. Factorials count the total arrangements of n distinct objects. The special case 0! = 1 is defined by mathematical convention. The intuitive reason: there is exactly one way to arrange zero objects — the empty arrangement. Formally, it follows from the recursive definition n! = n × (n−1)! applied at n=1: 1! = 1 × 0! → 0! = 1! / 1 = 1. Without this definition, the permutation and combination formulas would be undefined when r = 0 or r = n.

Use permutations when: assigning positions (1st, 2nd, 3rd place), creating ordered codes (PINs, passwords), forming anagrams, or any scenario where "who goes where" matters. Use combinations when: choosing a committee (no titles), selecting lottery numbers, dealing cards, or forming subsets where all members have equal standing. The keyword test: permutation problems use words like "arrange," "rank," "order," "sequence," "assign." Combination problems use "choose," "select," "group," "committee," "subset."

C(n,r) — read as "n choose r" — is the combination formula that calculates how many ways you can select r items from a total pool of n items when order does not matter. The formula is C(n,r) = n! / (r! × (n−r)!). For example, C(10,3) = 10! / (3! × 7!) = 120 tells you there are 120 ways to choose any 3 people from a group of 10. It equals the permutation P(10,3) = 720 divided by 3! = 6, because each group of 3 people can be arranged in 6 orders — all representing the same combination.

MISSISSIPPI has 11 letters: 1 M, 4 I's, 4 S's, and 2 P's. Using the formula for permutations with identical items: Total = 11! / (1! × 4! × 4! × 2!) = 39,916,800 / (1 × 24 × 24 × 2) = 39,916,800 / 1,152 = 34,650 distinct arrangements. The denominator divides out the duplicate orderings created by the repeated letters — swapping two I's, for example, gives the same word.

A standard "combination lock" is mathematically a permutation lock. The code 15-30-45 only opens the lock in that specific order — entering 30-15-45 leaves it locked. Because the sequence of numbers matters and creates a different result, this is a permutation by definition. A true combination would mean the lock opened for any order of those three numbers. The name "combination lock" is a historical misnomer that has confused students for generations — and is now one of the most effective mnemonics for remembering the permutation vs. combination distinction.

Counting methods determine both parts of the basic probability formula: P(A) = favorable outcomes / total outcomes. You use counting methods to calculate total outcomes (the size of the sample space) and to count how many of those outcomes satisfy the event. For example, to find P(dealt a pair in 5-card poker), you use C(52,5) = 2,598,960 for total hands and a combination calculation for "exactly one pair" hands to find the favorable count. Without counting methods, probability calculations are limited to very simple experiments.

A standard license plate format of 3 letters followed by 3 digits, with repetition allowed, uses the Fundamental Counting Principle: 26 choices per letter × 10 choices per digit = 26 × 26 × 26 × 10 × 10 × 10 = 26³ × 10³ = 17,576 × 1,000 = 17,576,000 possible plates. If repetition were not allowed: 26 × 25 × 24 × 10 × 9 × 8 = 11,232,000 plates. Different states use different formats (more digits, alphanumeric mixing), so always identify the exact format before calculating.

Authoritative External Resources

Open Textbook

OpenStax Discrete Mathematics

Free, peer-reviewed textbook covering combinatorics, permutations, and combinations with full derivations and exercises. Widely adopted in university discrete math courses.

Read on OpenStax →

University Courseware

MIT 6.042J — Mathematics for Computer Science

MIT OpenCourseWare lecture notes and problem sets. Counting and combinatorics are a core unit, presented as foundation for probability and algorithm analysis.

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Reference

Wolfram MathWorld — Combinatorics

Comprehensive mathematical reference for factorial, permutations, combinations, binomial coefficients, and multinomial distributions with formal notation and proofs.

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Free Textbook

Grinstead & Snell — Introduction to Probability (Dartmouth)

Free probability textbook from Dartmouth College. Chapter 3 covers combinatorics and counting methods in the context of probability theory with rigorous worked examples.

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Counting methods sit at the center of a broader mathematical framework. The pages below cover the topics that directly connect to what you have learned here on Statistics Fundamentals:

Prerequisite →