Fundamental counting principle, permutations, combinations, factorial, sample space, classical probability, addition and multiplication rules, conditional probability, expected value, and binomial probability — with full worked examples.
Permutations
P(n, r) = n! / (n − r)!
ordered selections, no repetition
Combinations
C(n, r) = n! / [r! · (n − r)!]
unordered selections, no repetition
Classical Probability
P(E) = n(E) / n(S)
equally likely outcomes
Addition Rule
P(A or B) = P(A) + P(B) − P(A and B)
subtract overlap
Multiplication Rule
P(A and B) = P(A) · P(B)
independent events only
Binomial Probability
P(X=k) = C(n,k) · p^k · (1−p)^(n−k)
n trials, k successes, prob p
The factorial of a non-negative integer n, written n!, is the product of all positive integers from 1 to n. It counts the number of ways to arrange n distinct objects in a row.
Definition
n! = n × (n − 1) × (n − 2) × … × 2 × 1
0! = 1 (by definition)
1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120
The special case 0! = 1 is not arbitrary. There is exactly one way to arrange zero objects (do nothing), and this definition keeps the recurrence n! = n × (n−1)! consistent when n equals 1.
Worked Example — Simplify a factorial ratio
Simplify 8! / 5!
Write out the cancellation:
8! / 5! = (8 × 7 × 6 × 5!) / 5! = 8 × 7 × 6 = 336
Key insight: cancel the common factorial in numerator and denominator rather than computing each factorial in full.
Worked Example — Large factorial computation
Evaluate 12! / (9! × 3!)
= (12 × 11 × 10 × 9!) / (9! × 6)
= (12 × 11 × 10) / 6
= 1320 / 6
= 220
This is C(12, 3) = 220, the number of ways to choose 3 items from 12.
The Fundamental Counting Principle (also called the multiplication rule for counting) states: if one task can be done in m ways and a second independent task can be done in n ways, then together the two tasks can be done in m × n ways. This extends to any number of independent choices.
General Statement
If task 1 has n₁ outcomes, task 2 has n₂ outcomes, …, and task k has nₖ outcomes, and all tasks are independent, then the total number of combined outcomes is
Total = n₁ × n₂ × … × nₖ
Worked Example — License Plates
A license plate consists of 3 letters followed by 4 digits. Letters and digits may repeat. How many distinct plates are possible?
Letters: 26 choices each → 26 × 26 × 26 = 17,576
Digits: 10 choices each → 10 × 10 × 10 × 10 = 10,000
Total: 17,576 × 10,000 = 175,760,000
Multiply across all independent stages; repetition is allowed so each stage has its full set of choices.
Worked Example — Outfit Combinations
You have 5 shirts, 4 pairs of pants, and 3 pairs of shoes. How many distinct outfits can you create?
5 × 4 × 3 = 60 outfits
Each choice is independent, so multiply the number of options at each stage.
A permutation is an ordered arrangement of objects. When you select r objects from n distinct objects and the order matters, you are counting permutations.
Formula — Permutations Without Repetition
P(n, r) = n! / (n − r)!
n = total objects available, r = objects chosen, order matters, no repetition allowed.
Why this formula? You have n choices for the first position, (n − 1) for the second, (n − 2) for the third, …, and (n − r + 1) for the r-th position. Multiplying these together gives n! / (n − r)!.
Worked Example — Race Finishers
Eight runners compete. In how many ways can the gold, silver, and bronze medals be awarded?
P(8, 3) = 8! / (8 − 3)!
= 8! / 5!
= 8 × 7 × 6
= 336
Order matters (gold ≠ silver), and no runner can win two medals, so this is a permutation without repetition.
Permutations With Repetition
When repetition is allowed, each of the r positions has n choices independently, so the count is simply:
n^r
Worked Example — PIN Code
A 4-digit PIN uses digits 0–9 with repetition allowed. How many PINs exist?
10^4 = 10,000
Each of the 4 positions independently has 10 choices.
Permutations of Repeated Objects
If n objects include groups of identical items — n₁ alike of one kind, n₂ alike of another, … — the number of distinct arrangements is:
n! / (n₁! × n₂! × …)
Worked Example — Arranging Letters
How many distinct arrangements of the letters in MISSISSIPPI are there?
M: 1, I: 4, S: 4, P: 2 (total 11 letters)
Arrangements = 11! / (1! × 4! × 4! × 2!)
= 39,916,800 / (1 × 24 × 24 × 2)
= 39,916,800 / 1,152
= 34,650
Dividing by the factorials of repeated counts removes duplicate arrangements that look identical due to swapping identical letters.
A combination counts the number of ways to choose r objects from n distinct objects when order does not matter. Combinations are used whenever selecting a group or committee, not an ordered ranking.
Formula
C(n, r) = n! / [r! × (n − r)!]
Also written as nCr, or “n choose r”, or the binomial coefficient (n over r).
Relationship to permutations: Every combination of r objects corresponds to r! permutations (all orderings of those same objects). Therefore C(n, r) = P(n, r) / r!.
| Scenario | Order Matters? | Formula |
|---|---|---|
| Choosing a committee of 3 from 10 people | No | C(10, 3) = 120 |
| Awarding 1st, 2nd, 3rd place to 10 runners | Yes | P(10, 3) = 720 |
| Selecting 5 cards from a 52-card deck | No | C(52, 5) = 2,598,960 |
| Arranging 4 books on a shelf | Yes | P(4, 4) = 4! = 24 |
Worked Example — Lottery
A lottery ticket requires choosing 6 numbers from 1 to 49. How many distinct tickets are possible?
C(49, 6) = 49! / [6! × 43!]
= (49 × 48 × 47 × 46 × 45 × 44) / (6 × 5 × 4 × 3 × 2 × 1)
= 10,068,347,520 / 720
= 13,983,816
Order of the chosen numbers does not matter (the ticket is the same regardless of draw order), so this is a combination.
Worked Example — Committee with Roles
From 12 students, how many ways can you select a committee of 4 with a designated chair, while the remaining 3 are ordinary members?
Step 1: Choose chair from 12 → 12 ways
Step 2: Choose 3 members from remaining 11 → C(11, 3) = 165 ways
Total: 12 × 165 = 1,980
The chair role is ordered (one specific person holds it), so we pick it first, then use a combination for the unordered ordinary members.
Before computing probabilities, we define the objects of study precisely.
Experiment
Any process that produces a definite outcome (rolling a die, drawing a card, flipping a coin).
Sample Space S
The set of ALL possible outcomes of an experiment. Rolling one die: S = {'{'}1, 2, 3, 4, 5, 6{'}'}.
Event E
Any subset of the sample space. Rolling an even number: E = {'{'}2, 4, 6{'}'}. Events are sets.
Complement of an Event
The complement of event E, written E′ (or E-complement or “not E”), is the set of all outcomes in S that are NOT in E. Together E and E′ partition the sample space.
P(E′) = 1 − P(E)
Worked Example — Two Dice
Two fair dice are rolled. Describe the sample space and find the number of outcomes in the event “the sum is 7.”
n(S) = 6 × 6 = 36 (ordered pairs (die1, die2))
Outcomes summing to 7:
(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
n(E) = 6
The sample space for two dice always has 36 equally likely outcomes when both dice are fair.
When all outcomes in the sample space are equally likely, the probability of event E is the ratio of favorable outcomes to total outcomes.
Classical Probability Formula
P(E) = n(E) / n(S)
n(E) = number of outcomes in E; n(S) = total outcomes in sample space S. Always: 0 ≤ P(E) ≤ 1.
Impossible Event
P(E) = 0
The event cannot occur (rolling a 7 on one standard die).
Certain Event
P(S) = 1
The event always occurs (rolling a number 1 through 6).
Worked Example — Card Draw
One card is drawn at random from a standard 52-card deck. Find the probability that the card is a heart.
n(S) = 52
n(heart) = 13
P(heart) = 13 / 52 = 1 / 4 = 0.25
There are 13 hearts in a 52-card deck.
Worked Example — Complementary Probability in Practice
A bag contains 3 red, 5 blue, and 2 green marbles. One marble is drawn at random. Find the probability it is NOT blue.
P(blue) = 5 / 10 = 1 / 2
P(not blue) = 1 − 1/2 = 1/2
Alternatively, P(not blue) = (3 + 2) / 10 = 5/10 = 1/2. Both approaches agree.
The addition rule gives the probability that at least one of two events occurs. Without adjustment, adding P(A) and P(B) counts any outcome in both A and B twice, so we subtract the overlap.
General Addition Rule
P(A or B) = P(A) + P(B) − P(A and B)
Mutually Exclusive Events
Events A and B are mutually exclusive if they cannot both occur: P(A and B) = 0. The addition rule simplifies to:
P(A or B) = P(A) + P(B) (mutually exclusive only)
Worked Example — Deck of Cards (Overlapping)
One card is drawn. Find the probability it is a king OR a heart.
P(king) = 4/52
P(heart) = 13/52
P(king AND heart) = 1/52 (king of hearts)
P(king OR heart) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13
The king of hearts is in both sets, so subtract it once to avoid double-counting.
Worked Example — Single Die (Mutually Exclusive)
A fair die is rolled. Find the probability of rolling a 2 or a 5.
P(2) = 1/6, P(5) = 1/6
Cannot roll both on one die → mutually exclusive
P(2 or 5) = 1/6 + 1/6 = 2/6 = 1/3
The multiplication rule gives the probability that two events both occur. The rule depends on whether the events are independent.
Independent Events
Events A and B are independent when the occurrence of one does not affect the probability of the other.
P(A and B) = P(A) × P(B) (independent)
Dependent Events
When the events are not independent, use the general multiplication rule involving conditional probability:
P(A and B) = P(A) × P(B|A)
Worked Example — Two Coin Flips (Independent)
A fair coin is flipped twice. Find the probability of getting heads both times.
P(H on flip 1) = 1/2
P(H on flip 2) = 1/2 (independent of flip 1)
P(both heads) = 1/2 × 1/2 = 1/4
Worked Example — Drawing Without Replacement (Dependent)
A bag has 4 red and 6 blue marbles. Two marbles are drawn without replacement. Find the probability both are red.
P(1st is red) = 4/10
P(2nd is red | 1st was red) = 3/9 (one red removed)
P(both red) = 4/10 × 3/9 = 12/90 = 2/15
After removing one red marble, only 3 red remain out of 9 total, making the second draw dependent on the first.
Worked Example — At Least One
A fair die is rolled 3 times. Find the probability of getting at least one six.
Use complement: P(at least one 6) = 1 − P(no 6 in 3 rolls)
P(no 6 on one roll) = 5/6
P(no 6 in 3 rolls) = (5/6)^3 = 125/216
P(at least one 6) = 1 − 125/216 = 91/216 ≈ 0.421
Complementary counting with the multiplication rule is far easier than enumerating all ways to get exactly one, exactly two, or exactly three sixes separately.
Conditional probability is the probability of event A given that event B has already occurred. Knowing B restricts the sample space to outcomes inside B.
Formula
P(A | B) = P(A and B) / P(B), provided P(B) ≠ 0
Read “P(A | B)” as “probability of A given B.”
Worked Example — Card Draw
One card is drawn from a standard deck. Given that the card is a face card (J, Q, or K), what is the probability it is a King?
There are 12 face cards (3 per suit × 4 suits).
There are 4 Kings.
All Kings are face cards, so P(King and face) = 4/52.
P(face card) = 12/52
P(King | face) = (4/52) / (12/52) = 4/12 = 1/3
Restricting to the 12 face cards, exactly 4 are Kings: 4/12 = 1/3.
Worked Example — Test for Independence via Conditional Probability
A and B are events with P(A) = 0.4, P(B) = 0.3, and P(A and B) = 0.12. Are A and B independent?
P(A | B) = P(A and B) / P(B) = 0.12 / 0.30 = 0.40 = P(A)
Since P(A | B) = P(A), events A and B are independent.
Also check: P(A) × P(B) = 0.4 × 0.3 = 0.12 = P(A and B). Both criteria confirm independence.
The complementary probability rule states that the probability of an event NOT occurring equals 1 minus the probability that it does occur. This is frequently the most efficient approach for “at least one” problems.
Rule
P(E′) = 1 − P(E)
Worked Example — Birthday Problem Setup
In a group of 3 people, what is the probability that at least two share a birthday? (Assume 365 equally likely birthdays, ignoring leap years.)
P(all different) = 365/365 × 364/365 × 363/365
= (365 × 364 × 363) / 365^3
≈ 0.9918
P(at least two share) = 1 − 0.9918 ≈ 0.0082
With only 3 people the probability is low, but it rises sharply with group size — surpassing 50% at just 23 people.
Worked Example — Quality Control
A factory produces parts where 2% are defective. If 5 parts are selected at random, find the probability that at least one is defective.
P(one part is not defective) = 0.98
P(none defective in 5) = 0.98^5 ≈ 0.9039
P(at least one defective) = 1 − 0.9039 ≈ 0.0961
About a 9.6% chance that the batch of 5 contains at least one defective part.
The expected value E(X) of a random variable X is the weighted average of all possible outcomes, where the weights are the probabilities of each outcome. It represents the long-run average over a large number of repetitions of the experiment.
Formula
E(X) = x₁ · P(x₁) + x₂ · P(x₂) + … + xₙ · P(xₙ)
E(X) = Σ xᵢ · P(xᵢ)
Worked Example — Fair Die
Find the expected value of one roll of a fair six-sided die.
E(X) = 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6)
= (1 + 2 + 3 + 4 + 5 + 6) / 6
= 21 / 6
= 3.5
3.5 is not a face on the die; it is the long-run average result over many rolls.
Worked Example — Carnival Game
A carnival game costs $2 to play. You roll one die: roll 6, win $10; roll 4 or 5, win $3; roll 1–3, win nothing. Find the expected net gain per play.
Net outcomes (after $2 entry fee):
Roll 6: +$8, P = 1/6
Roll 4 or 5: +$1, P = 2/6
Roll 1–3: −$2, P = 3/6
E = 8(1/6) + 1(2/6) + (−2)(3/6)
= 8/6 + 2/6 − 6/6
= 4/6
= $0.67 per play
A positive expected value means the player gains on average — this is a player-favorable game. Real casino games always have a negative expected value for the player.
Worked Example — Insurance Decision
An insurance policy costs $500/year. It pays $50,000 if a certain event occurs. The annual probability of the event is 0.005. Find the expected value of buying the policy.
Gain if event occurs: $50,000 − $500 = +$49,500, P = 0.005
Gain if no event: −$500, P = 0.995
E = 49,500(0.005) + (−500)(0.995)
= 247.50 − 497.50
= −$250
Expected value is −$250: you lose on average. Yet people rationally buy insurance because the risk reduction has personal utility beyond expected value alone.
Binomial probability applies when an experiment consists of n independent, identical trials, each with exactly two possible outcomes (success or failure), and the probability of success p is the same on every trial.
Conditions for a Binomial Experiment
Binomial Probability Formula
P(X = k) = C(n, k) · p^k · (1 − p)^(n − k)
C(n, k)
Number of ways to place k successes among n trials
p^k
Probability of the k successes occurring
(1 − p)^(n − k)
Probability of the (n − k) failures occurring
Worked Example — Coin Flips
A fair coin is flipped 5 times. Find the probability of getting exactly 3 heads.
n = 5, k = 3, p = 0.5
P(X = 3) = C(5, 3) · (0.5)^3 · (0.5)^2
= 10 · 0.125 · 0.25
= 10 · 0.03125
= 0.3125
C(5, 3) = 10 because there are 10 different sequences of 5 flips containing exactly 3 heads (e.g., HHHTT, HHTHT, …).
Worked Example — Free Throws
A basketball player makes free throws with probability 0.75. She shoots 8 free throws. Find the probability she makes exactly 6.
n = 8, k = 6, p = 0.75, q = 0.25
C(8, 6) = 8! / (6! × 2!) = 28
P(X = 6) = 28 · (0.75)^6 · (0.25)^2
= 28 · 0.17798 · 0.0625
≈ 28 · 0.011124
≈ 0.3115
About a 31.2% chance of making exactly 6 of 8 free throws.
Worked Example — At Least k Successes
The same player shoots 8 free throws (p = 0.75). Find the probability she makes at least 7.
P(X ≥ 7) = P(X = 7) + P(X = 8)
P(X = 7): C(8,7) · (0.75)^7 · (0.25)^1
= 8 · 0.13348 · 0.25 ≈ 0.2670
P(X = 8): C(8,8) · (0.75)^8 · (0.25)^0
= 1 · 0.10011 · 1 ≈ 0.1001
P(X ≥ 7) ≈ 0.2670 + 0.1001 = 0.3671
For “at least” questions, either sum from the cutoff up or use the complement.
Expected Value and Variance of a Binomial Distribution
Expected Value (Mean)
E(X) = n · p
In 8 free throw shots at 75%: E(X) = 8 × 0.75 = 6 makes on average.
Standard Deviation
σ = √(n · p · (1 − p))
In the free throw example: σ = √(8 × 0.75 × 0.25) = √1.5 ≈ 1.22
Problem 1 — Counting Paths
A password consists of 2 letters (A–Z) followed by 3 digits (0–9), with no letter repeated but digits may repeat. How many passwords are possible?
Solution
Letters (no repeat): P(26, 2) = 26 × 25 = 650
Digits (repeat allowed): 10^3 = 1,000
Total: 650 × 1,000 = 650,000
Problem 2 — Mixed Counting and Probability
A class has 10 boys and 8 girls. A committee of 5 is chosen at random. Find the probability the committee has exactly 3 girls.
Solution
Ways to choose 3 girls from 8: C(8, 3) = 56
Ways to choose 2 boys from 10: C(10, 2) = 45
Favorable: 56 × 45 = 2,520
Total committees from 18: C(18, 5) = 8,568
P(exactly 3 girls) = 2,520 / 8,568 = 315 / 1,071 ≈ 0.294
Problem 3 — Conditional Probability with a Table
In a survey of 200 students, 120 study math, 90 study science, and 50 study both. A student is chosen at random. Given that the student studies science, find the probability they also study math.
Solution
P(math and science) = 50 / 200 = 0.25
P(science) = 90 / 200 = 0.45
P(math | science) = 0.25 / 0.45 = 50/90 = 5/9 ≈ 0.556
Problem 4 — Expected Value with Multiple Prizes
A raffle sells 500 tickets at $5 each. Prizes: one $1,000 prize, two $200 prizes, five $50 prizes. Find the expected net gain for a single ticket.
Solution
Net gain if win $1,000: $995, P = 1/500
Net gain if win $200: $195, P = 2/500
Net gain if win $50: $45, P = 5/500
Net gain if win nothing: −$5, P = 492/500
E = 995(1/500) + 195(2/500) + 45(5/500) + (−5)(492/500)
= 995/500 + 390/500 + 225/500 − 2460/500
= (995 + 390 + 225 − 2460) / 500
= −850 / 500
= −$1.70
On average, each ticket loses $1.70. The organizer keeps $1.70 per ticket sold.
Problem 5 — Binomial with Complement
A multiple-choice exam has 10 questions, each with 4 choices. A student guesses randomly on every question. Find the probability of passing (at least 6 correct).
Solution
n = 10, p = 0.25, q = 0.75
P(X ≥ 6) = P(6) + P(7) + P(8) + P(9) + P(10)
P(6) = C(10,6)(0.25)^6(0.75)^4 = 210 × 0.000244 × 0.3164 ≈ 0.01622
P(7) = C(10,7)(0.25)^7(0.75)^3 = 120 × 0.0000610 × 0.4219 ≈ 0.00309
P(8) = C(10,8)(0.25)^8(0.75)^2 = 45 × 0.0000153 × 0.5625 ≈ 0.000386
P(9) = C(10,9)(0.25)^9(0.75)^1 = 10 × 0.00000381 × 0.75 ≈ 0.0000286
P(10) = C(10,10)(0.25)^10 ≈ 0.000000954
P(X ≥ 6) ≈ 0.01622 + 0.00309 + 0.000386 + 0.0000286 + 0.00000095
≈ 0.0197 ≈ 1.97%
Pure guessing yields only about a 2% chance of passing, as expected.
Permutation vs. Combination: The One-Question Test
Ask: “Does swapping two selected items create a different outcome?” If yes, use a permutation (P). If no, use a combination (C). Awarding gold/silver medals — swapping the two athletes changes who wins what, so P. Choosing a committee — swapping two members leaves the same committee, so C.
Mutually Exclusive vs. Independent
Mutually exclusive events cannot occur simultaneously: P(A and B) = 0. Independent events can both occur, but knowing one occurred does not change the other's probability. Mutually exclusive events with positive probabilities are always dependent (knowing A occurred tells you B did not).
Addition Rule vs. Multiplication Rule
Use the addition rule for “A OR B” (at least one occurs). Use the multiplication rule for “A AND B” (both occur). “Or” means union; “and” means intersection. Getting these confused is the most common probability error on precalculus exams.
When to Use Binomial vs. Classical Probability
Use binomial probability when you have repeated independent trials with the same success probability p. Use classical probability (n(E)/n(S)) for a single experiment with equally likely outcomes. Drawing one card calls for classical probability; flipping a coin 10 times and counting heads calls for binomial.
Error: Adding probabilities without subtracting the overlap
Fix: Always apply the full addition rule: P(A or B) = P(A) + P(B) − P(A and B). Only skip the subtraction when you know the events are mutually exclusive.
Error: Using P(A and B) = P(A) × P(B) for dependent events
Fix: The simple multiplication rule only holds for independent events. For dependent events, use P(A and B) = P(A) × P(B|A).
Error: Forgetting to account for repeated objects in permutations
Fix: When some objects are identical, divide n! by the factorial of each repetition count.
Error: Confusing P(A|B) with P(B|A)
Fix: These are generally not equal. P(King | face card) = 1/3, but P(face card | King) = 1. Always be precise about which event is the condition.
Error: Computing 0! = 0
Fix: 0! = 1 by definition. This is essential for the formulas C(n, 0) = C(n, n) = 1 to work correctly.
Counting Principles
Fundamental counting principle, tree diagrams, multiplication rule
Permutations and Combinations
P(n,r), C(n,r), factorial notation, with and without repetition
Probability
Sample space, events, classical probability, equally likely outcomes
Binomial Probability
Binomial distribution, P(X=k) formula, expected value of binomial
Expected Value
Weighted averages, fair games, decision analysis
Chapter 13 Review
Mixed counting and probability, all rules integrated
A permutation counts ordered arrangements (order matters); a combination counts unordered selections (order does not matter). For the same n and r, P(n, r) = r! × C(n, r) because each combination corresponds to r! permutations.
0! is defined as 1 so that formulas like C(n, 0) = n! / (0! × n!) = 1 are consistent. There is exactly one way to arrange zero objects, and the factorial recurrence 1! = 1 × 0! requires 0! = 1.
Use the addition rule whenever you need the probability that at least one of two events occurs (A OR B). Subtract P(A and B) to avoid double-counting outcomes in both events. If A and B are mutually exclusive, P(A and B) = 0 and the rule becomes P(A or B) = P(A) + P(B).
Two events are independent when knowing one occurred does not change the probability of the other: P(A|B) = P(A). For independent events, P(A and B) = P(A) × P(B). For dependent events, use P(A and B) = P(A) × P(B|A).
P(X = k) = C(n, k) × p^k × (1−p)^(n−k). The factor C(n, k) counts the number of arrangements of k successes in n trials. The factor p^k is the probability of those k successes, and (1−p)^(n−k) is the probability of the (n−k) failures.
P(A|B) is the probability A occurs given B has occurred. Formula: P(A|B) = P(A and B) / P(B). Knowing B restricts the sample space to outcomes inside B, so you scale by dividing by P(B).
Expected value E(X) = Σ x·P(x) gives the long-run average of a random variable. It is used to evaluate fairness of games, make insurance decisions, compare risky choices, and find the mean of a probability distribution.
Binomial Theorem
Pascal's Triangle and expanding (a+b)^n
Sequences & Series
Arithmetic, geometric, and summation notation
Combinations & Permutations
Deep dive into counting techniques
Statistics & Probability
Descriptive statistics and probability distributions
Mathematical Induction
Proof technique for counting identities
Precalculus Study Guide
Complete Stewart Precalculus chapter map
Counting problems and probability calculations require repeated practice to master. Work through adaptive problems that target permutations vs. combinations, conditional probability, and binomial applications — all aligned to Stewart Precalculus Chapter 13.
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