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Mathematics in Bukit Timah | Generating Functions, Coefficients and Recursive Counting

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by

eduKate Asia
Mathematics textbooks stand behind an open notebook containing written working and a graph, with a calculator and pens alongside.

Did you know that one algebraic expression can contain an entire infinite list of counting answers? The rational expression 1/(1−x−x²) expands as 1+x+2x²+3x³+5x⁴+8x⁵+⋯. Those coefficients count how many ways a strip of a given length can be tiled with pieces of length one and two when their order matters. The expression is a generating function, a bridge between sequences, algebra and combinatorics.

This Mathematics in Bukit Timah guide explains coefficient extraction, formal power series, geometric-series factors, convolution, unordered coins, ordered tilings, Fibonacci recurrences, binomial coefficients, bounded selections, integer partitions and probability generating functions. Every coin, tile, box, classroom choice and numerical record is fictional. No result reports actual Bukit Timah pupils, tuition enrolments, local shop transactions or measured events. The Mathematics counts arrangements under precisely stated assumptions.

Quick answer: The ordinary generating function for a sequence a₀,a₁,a₂,… is G(x)=Σ from n=0 to infinity aₙxⁿ. The notation [xⁿ]G(x)=aₙ means “take the coefficient of xⁿ.” For ordered tilings by lengths one and two, G(x)=1/(1−x−x²), giving [x⁵]G=8. For unordered combinations of unlimited one-unit and two-unit coins, the different expression 1/[(1−x)(1−x²)] has coefficient three at x⁵. Different rules produce different answers.

Article contents

Original eduKateSG Mathematics notebook and graph illustration accompanying fictional generating functions and combinatorial counting tasks.

Why generate a whole sequence instead of one count?

A counting problem may ask for the number of arrangements of length zero, one, two, three and so forth.

One could solve every size separately. Alternatively, place the size-n answer next to xⁿ in a single formal series.

The resulting algebra can reveal recurrences, factorisation rules and identities that are difficult to see from a short list of numbers.

The Bukit Timah context is educational, not observational

A learner near Beauty World, Sixth Avenue or King Albert Park can experiment with coloured tiles and abstract counting puzzles.

But the figures below are chosen to make counting transparent. They are not actual stock quantities, verified school data or local pricing rules.

Every coefficient describes the stated mathematical model, and no coefficient becomes a fact about local residents without appropriate evidence.

Start with one way at every size

Suppose there is exactly one permissible object for each size n≥0.

The sequence is 1,1,1,1,…, which we encode by 1+x+x²+x³+⋯.

As a formal power series it satisfies (1−x)(1+x+x²+⋯)=1, so it may be written 1/(1−x).

Worked example 1: verify the geometric-series identity

Multiply the series by one and then subtract x times the series.

The original series is 1+x+x²+x³+⋯. The shifted series is x+x²+x³+⋯.

Everything cancels except the constant term one. Therefore (1−x)G(x)=1, which explains the rational expression.

What does extracting a coefficient mean?

For polynomial H(x)=4+3x+7x²+2x³, the coefficient of x² is seven.

Write [x²]H(x)=7. Similarly [x⁰]H=4 and [x³]H=2.

The exponent identifies the size; the coefficient is the associated quantity. They must not be confused.

Worked example 2: a missing coefficient

Take P(x)=1+x³+x⁵.

There is no x² term, so [x²]P=0.

In a counting application, zero means no permitted arrangement of size two under the construction encoded by P.

A formal variable is a bookkeeping device

The symbol x in a combinatorial generating function is not automatically distance, money, time or a measured physical input.

One may manipulate it through coefficient identities without assigning a numerical value.

Substituting x=2 into an infinite geometric series would produce a divergent numerical sum, while extracting any fixed coefficient remains perfectly meaningful.

Count one optional item

A single fictional item of weight two has two possible statuses: omitted or selected.

Omission contributes 1, while selection contributes x².

Thus its selection polynomial is 1+x². The item may be selected at most once by the stipulated rule.

Worked example 3: combine two optional items

Let one item weigh two units and another weigh three.

The combined generating polynomial is (1+x²)(1+x³)=1+x²+x³+x⁵.

The coefficient at x⁵ is one, representing selecting both items. The coefficient at x¹ is zero because no permitted subset totals one.

Why multiplying factors combines choices

For two independent selection components, an outcome consists of one permitted choice from each component.

Their size markers multiply as xᵃxᵇ=xᵃ⁺ᵇ, recording the combined size.

If several pairs of choices give the same total, the resulting monomials collect into the same coefficient. That is how multiplication performs combinatorial counting.

Worked example 4: a coefficient of two

Expand (1+x+x²)(1+x²).

The product is 1+x+2x²+x³+x⁴.

The coefficient of x² is two because the target size can be formed by the first component contributing two and second zero, or first zero and second two.

Unlimited identical coins make a geometric factor

Imagine any nonnegative number of identical one-unit coins.

Possible contributions are 0,1,2,3,…, so the generating factor is 1+x+x²+x³+⋯=1/(1−x).

Each quantity has one selection. Reordering identical coins does not create a new combination.

Unlimited two-unit coins

An arbitrary nonnegative quantity of identical two-unit coins contributes total values 0,2,4,6,…

The factor is 1+x²+x⁴+⋯=1/(1−x²).

No odd total can be made with only those coins. The odd-power coefficients are therefore zero.

Combining one- and two-unit coins

With unlimited identical coins of values one and two, the combined generating function is

C(x)=1/[(1−x)(1−x²)].

The coefficient of xⁿ counts nonnegative integer pairs (a,b) satisfying a+2b=n, where a and b are the numbers of the two coin types.

Worked example 5: make six units

Solve a+2b=6 with a,b nonnegative integers.

Let b=0,1,2,3. The corresponding a values are 6,4,2,0.

There are four unordered combinations, so [x⁶]C(x)=4.

Worked example 6: make five units

For a+2b=5, b may be zero, one or two.

The associated a values are five, three and one.

Thus there are three unordered combinations. The sequence (1,2,2) is the same combination as (2,1,2) under an unordered-coin convention.

A direct formula confirms the coefficient

For a+2b=n, b can take any integer from zero through floor(n/2).

Each b uniquely determines a=n−2b, so the count is

floor(n/2)+1.

Thus the generating-function coefficient has an independent arithmetic proof.

Worked example 7: an amount of nine

For n=9, b may equal 0,1,2,3,4.

There are five choices, yielding floor(9/2)+1=5.

One can list the combinations directly, but the formula handles every nonnegative target n efficiently.

A third denomination adds another factor

With unlimited coins of values one, two and five, use

1/[(1−x)(1−x²)(1−x⁵)].

The coefficient [xⁿ] counts nonnegative solutions to a+2b+5c=n.

The same polynomial multiplication principle extends to any finite set of permitted denominations.

Worked example 8: target five with three denominations

For a+2b+5c=5, c is either zero or one.

If c=0, b can be 0,1,2: three choices. If c=1, then a=b=0: one choice.

Total four combinations. This count is distinct from ordered sequences of coin denominations.

Limited supplies require finite factors

Suppose at most two one-unit coins and at most one two-unit coin may be used.

The factors are 1+x+x² and 1+x², respectively.

Their product 1+x+2x²+x³+x⁴ encodes the bounded combinations. Using infinite geometric factors would incorrectly allow more coins than the problem permits.

Worked example 9: bounded target four

In the bounded problem, total four requires two one-unit coins and one two-unit coin.

No other permitted selection reaches four. Thus [x⁴]=1.

With unlimited one- and two-unit coins, total four would instead have three combinations: four ones, two ones and a two, or two twos.

Ordered tilings are a different counting problem

Consider a strip with length n, tiled left to right with pieces of length one or two.

The sequences (1,2) and (2,1) occupy the strip differently, so order matters.

Call the number of tilings Tₙ. The choice of the first tile creates a useful recurrence.

Derive the ordered tiling recurrence

If the first tile has length one, the remaining n−1 units can be filled in Tₙ₋₁ ways.

If the first tile has length two, the remaining n−2 units can be filled in Tₙ₋₂ ways.

These cases are disjoint and exhaustive, so Tₙ=Tₙ₋₁+Tₙ₋₂ for n≥2.

Why the zero-length case counts as one

A strip of length zero has one permissible empty tiling, so T₀=1.

A strip of length one has just the single one-unit tile, so T₁=1.

These initial values are not arbitrary. They support recurrence calculations and the algebraic generating function.

Worked example 10: enumerate length three

The ordered tilings of total length three are

(1,1,1), (1,2), and (2,1).

There are three, agreeing with T₃=T₂+T₁=2+1=3.

Worked example 11: enumerate length four

Length four has ordered tilings

(1,1,1,1), (1,1,2), (1,2,1), (2,1,1), and (2,2).

There are five. Their distinct left-to-right order is part of the mathematical definition.

Worked example 12: count length five

The recurrence gives T₂=2,T₃=3,T₄=5 and T₅=8.

Therefore a strip of length five has eight ordered tilings.

Contrast this with only three unordered ways to make five using one- and two-unit coins. Same sizes, different counted objects.

The tiling generating function

Write T(x)=T₀+T₁x+T₂x²+⋯.

Using the recurrence, all coefficients of degree n≥2 in (1−x−x²)T(x) vanish.

The constant coefficient is T₀=1 and the x coefficient is T₁−T₀=0, giving (1−x−x²)T(x)=1.

Worked example 13: solve the identity

Divide formally by 1−x−x², whose constant coefficient is one.

The generating function is T(x)=1/(1−x−x²).

Expanding produces 1+x+2x²+3x³+5x⁴+8x⁵+⋯, matching every directly counted small tiling value.

Fibonacci numbers appear with a shifted index

Define F₀=0,F₁=1 and Fₙ=Fₙ₋₁+Fₙ₋₂.

The Fibonacci generating function is F(x)=x/(1−x−x²).

Our tiling numbers satisfy Tₙ=Fₙ₊₁, because their starting values are T₀=T₁=1 rather than F₀=0,F₁=1.

Worked example 14: compare sequences

Tiling counts begin 1,1,2,3,5,8.

Fibonacci numbers begin 0,1,1,2,3,5,8.

The tiling count for n=5 is T₅=8, corresponding to Fibonacci F₆=8. Matching the recurrence does not eliminate the importance of initial values.

A recurrence can determine every coefficient

Once T₀ and T₁ are known, T₂, T₃ and every later term follow by addition.

The rational generating function stores the initial conditions in its numerator and the recurrence in its denominator.

This provides a compact algebraic representation of an infinite sequence rather than merely a memory aid for a short pattern.

A series for the sequence one, two, three, four

Consider aₙ=n+1 for n≥0.

Its generating function is 1+2x+3x²+4x³+⋯=1/(1−x)².

One derivation formally differentiates the geometric series 1+x+x²+⋯ to obtain 1+2x+3x²+⋯, then shifts as needed.

Worked example 15: extract the sixth value

From G(x)=1/(1−x)², the coefficient of x⁵ is 6.

That matches the sequence rule a₅=5+1=6.

The denominator’s exponent two does not mean there are only two terms. It controls an infinite coefficient sequence.

Stars and bars comes from repeated factors

Suppose n identical tokens are placed into three labelled boxes A,B,C with no capacity limits.

Each box contributes 1/(1−x), so the combined generating function is 1/(1−x)³.

The coefficient of xⁿ is C(n+2,2), counting the nonnegative solutions of a+b+c=n.

Worked example 16: seven tokens into three boxes

For n=7, the number of labelled-box distributions is

C(7+2,2)=C(9,2)=9×8/2=36.

The boxes are distinct, while the tokens are identical. A different assumption about labels would require a different counting formula.

Binomial coefficients arise from optional choices

For m distinct optional items, each of weight one, each contributes factor (1+x).

Their combined polynomial is (1+x)ᵐ.

Choosing exactly k items contributes xᵏ, so [xᵏ](1+x)ᵐ=C(m,k). This provides a combinatorial explanation of the binomial expansion.

Worked example 17: choose three of five

There are five distinguishable items and each can be chosen once.

The coefficient of x³ in (1+x)⁵ is C(5,3)=10.

Each term corresponds to selecting x from three different factors and one from the other two factors.

Different weights produce subset-sum polynomials

Suppose three optional items have weights two, three and four.

Their combined polynomial is (1+x²)(1+x³)(1+x⁴).

Every term corresponds to a subset, and its exponent records total weight. Its coefficient tells how many different subsets achieve that total.

Worked example 18: subset total five

With weights two, three and four, the only subset summing to five is the pair {2,3}.

Therefore [x⁵](1+x²)(1+x³)(1+x⁴)=1.

No subset sums to one, so [x¹] is zero. Coefficients can encode impossibility as well as positive counts.

Multiplication produces a convolution sum

Let A(x)=Σaₙxⁿ and B(x)=Σbₙxⁿ.

Their product’s coefficient of xⁿ is

Σ from k=0 to n aₖbₙ₋ₖ.

This is called convolution. It adds the contributions from every possible division of total size n between the two components.

Worked example 19: calculate one coefficient

Let A=1+2x+x² and B=1+3x+2x².

At x², the product coefficient is

a₀b₂+a₁b₁+a₂b₀=1×2+2×3+1×1=9. This matches ordinary polynomial multiplication.

Why the constant term is important

The coefficient at x⁰ often represents the empty selection, which is a genuine combinatorial object under the stated convention.

For the product of two selection functions, the constant term is the product of their two constant terms.

Omitting the empty outcome can change a recurrence and every coefficient that depends on it.

Ordinary versus exponential generating functions

An ordinary generating function uses terms aₙxⁿ.

An exponential generating function instead uses aₙxⁿ/n!.

Exponential generating functions are particularly useful for certain labelled combinatorial structures. The two types encode the same sequence under different normalisations but do not have identical formulas.

Worked example 20: two functions for one sequence

For the constant sequence aₙ=1, the ordinary generating function is 1/(1−x).

The exponential generating function is eˣ, because eˣ=Σxⁿ/n!.

Both encode coefficient sequence one under their own conventions. One should not extract an ordinary coefficient from eˣ and call it aₙ without accounting for n!.

Probability generating functions use probabilities

For a nonnegative integer-valued random variable X, define

G_X(s)=Σ P(X=k)sᵏ, summing over k≥0.

The coefficients are probabilities and add to one. This is a different interpretation from counting the number of combinatorial arrangements.

Worked example 21: a Bernoulli variable

Suppose X is one with probability p and zero with probability 1−p.

Then G_X(s)=(1−p)+ps.

Substituting s=1 gives one. The constant coefficient is P(X=0), and the coefficient of s is P(X=1).

Independent random counts can be added by multiplication

If X and Y are independent nonnegative integer-valued random variables, then

G_{X+Y}(s)=G_X(s)G_Y(s).

The multiplication reflects adding independent random counts and summing over their possible paired outcomes.

If the variables are dependent, the product need not describe their sum correctly.

Worked example 22: three independent fair trials

The PGF of one fair binary trial is (1+s)/2.

For three independent trials, the sum has PGF [(1+s)/2]³=(1+3s+3s²+s³)/8.

The probabilities for zero, one, two and three successes are 1/8,3/8,3/8,1/8, which add to one.

Expected value from a PGF derivative

For a suitable PGF, G’_X(1)=E[X] when the expectation exists.

Differentiate [(1+s)/2]³ to obtain (3/2)[(1+s)/2]².

At s=1, the expected number of successes is 3/2=1.5, which need not be a possible realised integer count.

Worked example 23: check by direct expectation

For the three-trial distribution, E[X] is

0×1/8+1×3/8+2×3/8+3×1/8.

This equals (3+6+3)/8=12/8=1.5. The derivative method and direct probability average agree.

The geometric-series convergence warning

The formal identity 1/(1−x)=1+x+x²+⋯ is valid coefficient by coefficient.

As a numerical series, it converges only when |x|<1.

At x=2 the numerical sum 1+2+4+8+⋯ diverges, even though 1/(1−2)=−1 is defined. A rational expression and its local infinite expansion are not interchangeable at every numerical x.

Formal rational functions can encode recurrences

For a rational generating function G=P/Q whose denominator Q has nonzero constant coefficient, formal power-series division determines the coefficients.

Multiplying by Q often yields a linear recurrence with constant coefficients beyond certain initial terms.

For 1/(1−x−x²), coefficient comparison gives the Fibonacci recurrence directly.

Worked example 24: recover recurrence from coefficients

Let T(x)=ΣTₙxⁿ and suppose (1−x−x²)T(x)=1.

For n≥2, coefficient of xⁿ on the left is Tₙ−Tₙ₋₁−Tₙ₋₂.

Since the right side has zero coefficient at those powers, Tₙ=Tₙ₋₁+Tₙ₋₂. This is an exact formal calculation.

Integer partitions extend unlimited-coin thinking

An integer partition represents a nonnegative integer as an unordered sum of positive integer parts.

Allow parts of size one, two, three and so on, with unlimited repetitions.

The formal generating product is ∏ from j=1 to infinity 1/(1−xʲ). For a fixed coefficient xⁿ, factors with j>n cannot affect it.

Worked example 25: partitions of four

The unordered partitions of four are

4; 3+1; 2+2; 2+1+1; 1+1+1+1.

There are five, so the coefficient of x⁴ in the partition generating product equals five.

A composition is different from a partition

A composition of an integer is an ordered list of positive parts summing to that integer.

For four, the orders 3+1 and 1+3 are distinct compositions but the same unordered partition.

Thus the counting model—and its generating-function construction—changes when the order convention changes.

Worked example 26: count compositions of three

The compositions of three are

3; 2+1; 1+2; 1+1+1.

There are four, while the unordered partitions of three are 3,2+1,1+1+1, only three. The counts differ entirely because of order.

Matrix-valued generating functions count graph walks

If A is an adjacency matrix of a finite network, the entry (Aⁿ)ᵢⱼ counts n-edge walks from vertex i to j under the standard graph convention.

Formally, the matrix series I+xA+x²A²+⋯ is (I−xA)⁻¹ when interpreted as a formal matrix series around x=0.

This is an advanced extension showing that coefficient methods can encode network walks as well as coins and tiles.

Worked example 27: two steps on a short path

Consider graph vertices A—B—C.

There is exactly one two-edge walk from A to C: A→B→C.

Therefore the A,C entry of the adjacency matrix squared equals one, and the coefficient of x² in that matrix-series entry is one.

A generating function does not create real-world evidence

A mathematical coefficient is exact for its prescribed combinatorial rules.

It cannot establish how many real pupils will choose an activity, how many products a shop will sell, or which local service is available.

Those questions require actual data. Our numbers are deliberately fictional teaching results.

Modelling is the most important step

Before selecting factors, determine what the objects are, which sizes they contribute, whether repetition is permitted and whether order makes arrangements distinct.

Only then does the generating function reflect the intended question.

Algebraically expanding a beautifully written but incorrectly chosen product produces an exact answer to the wrong counting model.

Common error 1: confuse coefficient with exponent

In 5x⁴, the exponent four records the size index and the coefficient five records the quantity.

Thus [x⁴](5x⁴)=5.

The two numbers serve different roles and should not be interchanged.

Common error 2: count the same unordered coin arrangement repeatedly

The coin selections (1,2,2) and (2,1,2) use identical denomination counts.

If the model treats coins of each value as indistinguishable and order irrelevant, they are one combination.

Ordered tilings, in contrast, can count both as different arrangements.

Common error 3: omit supply limits

An item available at most once contributes 1+xʷ, not 1/(1−xʷ).

The unlimited factor creates selections with two, three or more copies that are forbidden in the bounded problem.

Match the generating factor to the supply rule.

Common error 4: forget the empty arrangement

In the tiling recurrence, T₀=1 counts the unique empty arrangement.

Setting T₀=0 without changing the model breaks the early recurrence and generating-function numerator.

Define the size-zero case explicitly before writing the series.

Common error 5: multiply dependent probability generating functions

For PGFs, multiplication describes independent sums under the stated model.

Without independence, the joint probabilities of X and Y cannot in general be recovered from their separate marginal PGFs alone.

Do not transfer a combinatorial product shortcut into an incompatible probability model.

Common error 6: treat formal series as numerical equalities everywhere

The series 1+x+x²+⋯ does not converge numerically at x=2.

Its formal relation with 1/(1−x) concerns coefficient arithmetic, not summing the divergent numerical series to minus one.

State the intended interpretation.

Common error 7: claim all coefficients are measured quantities

The coefficient eight for ordered length-five tilings counts mathematical possibilities, not observed events.

No real Bukit Timah school data were supplied to the article.

Separate combinatorial counting from statistical or geographic evidence.

Primary readiness: small selection and arrangement puzzles

Primary learners can make lists of coins, coloured pieces and ordered tilings for small totals.

They can discover that reversing two different tiles gives a different ordered arrangement but reversing identical coins does not create a new combination.

This understanding is more useful initially than memorising advanced formal power-series notation.

Lower Secondary: polynomials and sequences

Students can expand (1+x²)(1+x³), identify terms and build tiling recurrences.

Geometric sequences and binomial coefficients offer natural connections.

Full formal-series inversion can be introduced later when the student understands why the coefficients count valid configurations.

Advanced Mathematics and syllabus boundaries

Generating functions, formal power-series algebras and exponential generating functions are discrete Mathematics enrichment.

Consult the SEAB 2027 G3 Mathematics syllabus directory for actual school scope. The advanced techniques here are not claimed to be compulsory SEC G3 examination content.

The transferable principle is precise modelling of what is counted.

A twenty-minute parent diagnostic

Minutes 1–5: Count unordered one- and two-unit coin combinations making five, obtaining three.

Minutes 6–10: Count ordered one- and two-unit tilings of length five, obtaining eight via a recurrence.

Minutes 11–15: Write the different generating functions and explain why their denominators differ.

Minutes 16–20: Expand (1+x+x²)(1+x²), identify [x²]=2, and explain what finite supply restrictions mean.

Seven days of practice and transfer

Day one extracts coefficients. Day two builds optional-item factors. Day three counts unlimited coin combinations.

Day four derives ordered tiling recurrence. Day five manipulates its rational generating function. Day six practises convolution and binomial coefficients.

Day seven distinguishes counting from probability PGFs and checks model assumptions before multiplying.

Practice questions with checked answers

A. [x³](1+2x+3x²+4x³)? Answer: 4.

B. Unlimited one- and two-unit coins: ways to make six? Answer: 4.

C. Ordered tilings of length five? Answer: 8.

D. Tiling generating function? Answer: 1/(1−x−x²).

E. Fibonacci F₀=0,F₁=1 generating function? Answer: x/(1−x−x²).

F. [x³](1+x)⁵? Answer: C(5,3)=10.

G. Distribute seven identical tokens among three labelled boxes? Answer: C(9,2)=36.

H. [x²](1+x+x²)(1+x²)? Answer: 2.

Frequently asked questions

What is a generating function?

A formal or analytic power series whose coefficients encode an indexed sequence, often of combinatorial counts.

What does [xⁿ] mean?

It selects the coefficient multiplying xⁿ in the given power series.

Does x have to be assigned a numerical value?

No. In formal combinatorics it can be a bookkeeping symbol for object size.

Why does polynomial multiplication count combined choices?

Exponents add when selections combine, and coefficients add contributions from different ways to reach the same size.

Why do coin and tiling answers differ?

Coin combinations ignore order, whereas tilings count distinct left-to-right sequences.

What does the Fibonacci recurrence have to do with tilings?

Classifying by the first tile yields Tₙ=Tₙ₋₁+Tₙ₋₂, the Fibonacci recurrence with shifted starting values.

Are PGFs the same as ordinary counting generating functions?

They use similar algebra, but PGF coefficients are probabilities and require statistical assumptions.

Does a formal geometric series converge at every number?

No. Formal coefficient identities and numerical convergence are different concepts.

Are the counts from real Bukit Timah schools or shops?

No. All numerical objects and arrangements are invented teaching models.

How can parents recognise mastery?

The learner specifies whether order and repetition matter, writes suitable factors, and independently verifies small coefficients by enumeration.

Official sources and connected Mathematics in Bukit Timah

Consult the SEAB 2027 G3 Mathematics and Additional Mathematics syllabus directory for the appropriate school context. Generating-function theory here is enrichment, not an assertion of universal compulsory examination content.

Continue with Grid Paths, Pascal Triangle and Combinations, Binomial Expansions, Mathematical Induction and Recursion, Second Differences and Quadratic Sequences and Expected Value and Probability Distributions.

The immutable Clementi Secondary 1 Mathematics small-group tutorials reference anchors first-principles teaching, worked verification, misconception repair and independent practice.

The lasting habit is to define the counted objects before writing algebra, identify whether order and repetition matter, extract the correct coefficient and check the result against small cases.

Companion Mathematics in Bukit Timah lessons in this four-article batch: Pigeonhole Principle, Guaranteed Matches and Counting Proofs · Jensen’s Inequality, Convexity and Why Averages Behave Differently · Poisson Processes, Random Arrivals and Exponential Waiting Times. These new guides connect deterministic counting, generating functions, convex averages and stochastic processes, each with original fictional worked examples and carefully stated assumptions.