The Central Idea
A sequential choice is a choice made as one stage of a larger process.
If a complete outcome requires several stages, we count the available choices at each stage and combine them using the product principle.
If the stages have
\[ n_1,n_2,\ldots,n_k \]available choices, then the total number of complete outcomes is
\[ \boxed{n_1n_2\cdots n_k}. \]Think in Stages
Suppose a meal consists of:
- one entrée;
- one side;
- one beverage.
If there are \(5\) entrées, \(4\) sides, and \(3\) beverages, then the stages have
\[ 5,\qquad4,\qquad3 \]choices.
Therefore,
\[ \boxed{5\cdot4\cdot3=60}. \]One Complete Outcome Uses Every Required Stage
A useful first step is to identify what one complete outcome contains.
For the meal example, one outcome can be represented as
\[ (\text{entrée},\text{side},\text{beverage}). \]The ordered positions show the three required stages.
Counting complete outcomes means counting all valid combinations of choices across those stages.
Example: Building a Code
Suppose a code contains:
- one letter;
- one digit;
- one letter.
If all \(26\) letters and all \(10\) digits are allowed, with repetition permitted, then
\[ 26\cdot10\cdot26=6760. \]Therefore,
\[ \boxed{6760} \]codes are possible.
The Order of the Stages Matters
Sequential choices often occur in labeled positions.
A code of the form
\[ \text{letter-digit} \]is structurally different from a code of the form
\[ \text{digit-letter}. \]Even when the same symbols are used, their positions can produce different outcomes.
Repeated Choices
Sometimes the same option may be used more than once.
Suppose a four-digit code uses digits \(0\) through \(9\), and repetition is allowed.
Every stage has \(10\) choices:
\[ 10\cdot10\cdot10\cdot10=10^4. \]Therefore,
\[ \boxed{10{,}000} \]codes are possible.
When Repetition Is Not Allowed
If an option cannot be reused, the number of choices may decrease after each stage.
Suppose a three-letter string is formed from \(8\) distinct letters without repetition.
There are:
- \(8\) choices for the first position;
- \(7\) choices for the second position;
- \(6\) choices for the third position.
Thus,
\[ \boxed{8\cdot7\cdot6=336}. \]Earlier Choices Can Affect Later Choices
Sequential choices are not always independent.
If choosing an option at one stage removes or changes options at a later stage, the later count must reflect that change.
This is why the stage counts in a product need not be equal.
Example: Assigning Offices
Three different offices are assigned from a group of \(10\) people:
- president;
- vice president;
- secretary.
One person cannot hold two offices.
There are
\[ 10 \]choices for president,
\[ 9 \]remaining choices for vice president, and
\[ 8 \]remaining choices for secretary.
Therefore,
\[ \boxed{10\cdot9\cdot8=720}. \]Position Restrictions
Some restrictions apply only at particular stages.
For example, the first digit of a positive integer cannot be \(0\).
A four-digit positive integer therefore has:
- \(9\) choices for the first digit;
- \(10\) choices for each remaining digit.
Thus,
\[ \boxed{9\cdot10\cdot10\cdot10=9000}. \]Stage-Specific Restrictions
Suppose a five-character password must begin with a letter and end with a digit.
The middle three positions may contain either letters or digits.
Assuming \(26\) letters and \(10\) digits, with repetition allowed, the stage counts are
\[ 26,\qquad36,\qquad36,\qquad36,\qquad10. \]Therefore,
\[ \boxed{26\cdot36^3\cdot10}. \]Constant Choice Counts
If every stage has the same number \(n\) of choices and there are \(k\) stages, then
\[ \underbrace{n\cdot n\cdots n}_{k\text{ factors}}=n^k. \]This pattern appears frequently in strings, codes, functions, and state descriptions.
Binary Strings
A binary string uses the symbols \(0\) and \(1\).
Each position has \(2\) choices.
A binary string of length \(n\) therefore has
\[ \boxed{2^n} \]possible forms.
Example: Eight Binary Choices
A sequence contains \(8\) yes-or-no decisions.
Each decision has \(2\) possible outcomes.
\[ 2^8=256. \]Therefore,
\[ \boxed{256} \]complete decision sequences are possible.
Sequential Choices and Functions
A function from a finite domain can be specified by choosing an output for each input.
Suppose a domain has \(4\) elements and the codomain has \(5\) elements.
Each of the \(4\) domain elements can be assigned any of \(5\) outputs.
Therefore,
\[ \boxed{5^4=625} \]functions are possible.
Sequential Choices and Paths
A route can also be viewed as a sequence of choices.
Suppose a traveler must go from \(A\) to \(B\), then from \(B\) to \(C\), then from \(C\) to \(D\).
If there are \(3\), \(5\), and \(2\) route choices at those stages, then
\[ \boxed{3\cdot5\cdot2=30} \]complete routes are possible.
Sequential Choices and State Spaces
Many systems are described by several variables or settings.
Suppose a machine has:
- \(3\) speed settings;
- \(4\) temperature settings;
- \(2\) operating modes.
A complete machine state is determined by one choice from each category:
\[ \boxed{3\cdot4\cdot2=24}. \]Not Every Sequential Process Has a Single Uniform Product
Sometimes the number of available choices at a later stage depends on which earlier choice was made.
If different branches have different numbers of continuations, one simple product may no longer describe the entire process.
In that case, we can count the branches separately and add their totals.
Example: Different Numbers of Continuations
A traveler chooses one of two first-stage routes.
- Route \(A\) leads to \(4\) possible second-stage routes.
- Route \(B\) leads to \(7\) possible second-stage routes.
The total number of complete trips is
\[ 4+7=11. \]We cannot write \(2\cdot4\) or \(2\cdot7\) because the two first-stage choices do not have the same number of continuations.
Split into Cases When Necessary
A useful strategy is:
- identify different first-stage cases;
- count the continuations within each case;
- add the case totals.
This combines the product principle within branches and the sum principle across branches.
Example: Restricted Meal Choices
A restaurant has \(3\) entrées.
- Entrée \(A\) can be paired with \(4\) sides.
- Entrée \(B\) can be paired with \(5\) sides.
- Entrée \(C\) can be paired with \(2\) sides.
The total number of valid entrée-side meals is
\[ 4+5+2=11. \]This is a sequential process, but the later choice count varies by branch.
Choosing the Best Stage Order
When counting, the conceptual order of choices can sometimes be reorganized.
A useful stage order makes restrictions easy to track.
For example, when forming a number with no repeated digits, it is natural to choose the restricted first digit first, then count the remaining positions.
Example: Three-Digit Numbers without Repetition
Count three-digit positive integers with no repeated digits.
The first digit has \(9\) choices:
\[ 1,2,\ldots,9. \]Once it is chosen, the second digit has \(9\) remaining choices because \(0\) is now allowed but the first digit cannot repeat.
The third digit then has \(8\) choices.
\[ \boxed{9\cdot9\cdot8=648}. \]Tracking Remaining Options
In no-repetition problems, it is often helpful to think in terms of how many unused options remain.
If \(n\) distinct objects fill \(k\) ordered positions without repetition, the stage counts are
\[ n,\quad n-1,\quad n-2,\quad\ldots,\quad n-k+1. \]Their product is
\[ \boxed{n(n-1)(n-2)\cdots(n-k+1)}. \]Later counting lessons will connect this pattern to permutations.
Sequential Choices with Categories
Some processes require a fixed number of choices from different categories.
Suppose a product is configured by choosing:
- one of \(4\) sizes;
- one of \(6\) colors;
- one of \(3\) materials;
- one of \(2\) finishes.
The number of configurations is
\[ \boxed{4\cdot6\cdot3\cdot2=144}. \]Sequential Choices with Conditional Restrictions
Sometimes a restriction is activated only after a particular choice.
For example, a product may allow \(5\) color choices for one material but only \(3\) color choices for another.
The later stage then depends on the earlier material choice.
Such problems are often easier to represent with cases or decision trees.
Sequential Choices and Decision Trees
A decision tree draws each stage as a branching process.
Each complete path from the starting point to an endpoint represents one complete outcome.
When every branch at a stage has the same number of continuations, the tree visually demonstrates the product principle.
When branch sizes differ, the tree helps us see which case counts must be added.
Common Error: Using the Original Number at Every Stage
Suppose \(5\) different books are arranged in \(3\) ordered positions without repetition.
The incorrect count
\[ 5^3 \]allows the same book to be used more than once.
The correct stage counts are
\[ 5,\qquad4,\qquad3. \]Therefore,
\[ \boxed{5\cdot4\cdot3=60}. \]Common Error: Forgetting a Position Restriction
A four-digit positive integer cannot begin with \(0\).
Therefore,
\[ 10^4 \]is not the correct count.
The correct count is
\[ \boxed{9\cdot10^3=9000}. \]Common Error: Multiplying Unequal Branch Counts
Suppose one first-stage choice has \(3\) continuations and another has \(6\).
There is no single common continuation count.
The correct total is
\[ 3+6=9, \]not
\[ 2\cdot3 \]or
\[ 2\cdot6. \]Common Error: Counting Partial Outcomes
If a complete outcome requires three stages, counting only the possibilities from the first two stages does not solve the problem.
Every required stage must appear in the counting structure.
A Reliable Procedure
- Define one complete outcome.
- Break the outcome into sequential stages.
- Count the available choices at the first stage.
- Update the number of choices at later stages as restrictions take effect.
- Multiply along a branch when the stage counts are known.
- If different branches have different continuation counts, count them separately.
- Add the totals of disjoint branches.
- Check that every valid outcome appears exactly once.
Example: Full Sequential Workflow
Count four-character strings formed from the letters \(A,B,C,D,E\) with no repetition.
First position:
\[ 5 \]choices.
Second position:
\[ 4 \]remaining choices.
Third position:
\[ 3 \]remaining choices.
Fourth position:
\[ 2 \]remaining choices.
Therefore,
\[ \boxed{5\cdot4\cdot3\cdot2=120}. \]Practice 1: Three Stages
A meal uses one of \(4\) entrées, one of \(5\) sides, and one of \(3\) beverages. How many meals are possible?
Practice 2: Repetition Allowed
How many five-digit codes can be formed using digits \(0\) through \(9\) if repetition is allowed?
Practice 3: No Repetition
How many four-letter strings can be formed from \(9\) distinct letters if no letter repeats?
Practice 4: Position Restriction
How many five-digit positive integers are there?
Practice 5: Binary Choices
How many binary strings of length \(10\) are possible?
Practice 6: Functions
How many functions are there from a three-element set to a six-element set?
Practice 7: Assigning Roles
From \(12\) people, how many ways can a chair, vice chair, and secretary be assigned if no person may hold more than one role?
Practice 8: Mixed Character Types
How many four-character codes have the form letter-digit-digit-letter if repetition is allowed?
Practice 9: Unequal Branches
A first-stage choice has three options. The first option has \(2\) continuations, the second has \(5\), and the third has \(4\). How many complete outcomes are possible?
Practice 10: Three-Digit Numbers without Repetition
How many three-digit positive integers can be formed if no digit repeats?
Practice Answers
Practice 1
\[ \boxed{4\cdot5\cdot3=60}. \]Practice 2
\[ \boxed{10^5=100{,}000}. \]Practice 3
\[ \boxed{9\cdot8\cdot7\cdot6=3024}. \]Practice 4
The first digit has \(9\) choices and each remaining digit has \(10\):
\[ \boxed{9\cdot10^4=90{,}000}. \]Practice 5
\[ \boxed{2^{10}=1024}. \]Practice 6
Each of the \(3\) domain elements may be assigned any of \(6\) outputs:
\[ \boxed{6^3=216}. \]Practice 7
\[ \boxed{12\cdot11\cdot10=1320}. \]Practice 8
\[ \boxed{26\cdot10\cdot10\cdot26=67{,}600}. \]Practice 9
Add the continuation counts of the three disjoint first-stage branches:
\[ \boxed{2+5+4=11}. \]Practice 10
There are \(9\) choices for the first digit, \(9\) choices for the second, and \(8\) choices for the third:
\[ \boxed{9\cdot9\cdot8=648}. \]What to Remember
Sequential counting follows a simple organizing idea:
\[ \boxed{\text{complete outcomes} =(\text{choices at stage 1}) (\text{choices at stage 2})\cdots}. \]- Define what one complete outcome contains.
- Break the process into stages.
- Allow stage counts to change when restrictions apply.
- Use decreasing counts when choices cannot be repeated.
- Split into cases when different branches have different continuations.
- Count every valid outcome once and only once.
Where We Go Next
Sequential choices can be visualized as branches growing from one decision to the next.
The next lesson, Decision Trees, turns this stage-by-stage structure into a visual counting method that is especially useful when different choices lead to different numbers of later possibilities.