Tricks & Memory Hooks

IUVB — the four special relations

Identity (loops only) · Universal (whole grid) · Void (empty) · Binary (the umbrella that contains the others)

R-S-T — the three properties

Reflexive = Reflect, you see yourself → loops on everything

Symmetric = Symmetry, a mirror → every arrow has a twin

Transitive = Transit, you pass through b → the long arrow a→c is forced

Quick tests that beat checking pair by pair

• Reflexive, fast version: count elements in A, count loops in the diagram. Equal? Reflexive.
• Symmetric, fast version: a relation that is its own mirror is symmetric — check if the diagram looks the same flipped.
• Transitive, fast version: only look for chains of length 2 (a→b→c). If there are none, it is automatically transitive.

Cartesian product

"Outside × inside." Write the first set down the side, the second set across the top, and fill the grid. Answer = number of cells.

Floor vs ceiling — the cup and the hat

⌊  ⌋ is a cup (rounds down into it). ⌈  ⌉ is a hat (rounds up under it).

Negative-number trap: ⌊−1.5⌋ = −2, not −1. Check: the floor is always ≤ x, the ceiling always ≥ x. If your answer breaks that, redo it.

Exponential curve

Everything passes through (0, 1) because a⁰ = 1. So just ask: is a bigger or smaller than 1? Bigger = goes up, smaller = goes down. And it never touches the x-axis (y > 0 always).

Booleans

AND = multiplication (one zero kills it) · OR = addition (one is enough) · NOT = 1 − x · XOR = output 1 only when inputs differ.

Recursion

Down to the base case, then back up. Write the call stack in a column going down to smaller numbers, then compute the column up. Marks come for the base case even if the arithmetic slips.

Homogeneous recurrences

• Always verify your final closed form by substituting back into the original recurrence.
• Two distinct roots → A r₁ⁿ + B r₂ⁿ. Repeated root → (A + Bn) rⁿ.
• Quote the characteristic equation, not just the answer.

The master question

Does the order of the chosen things matter?

YES → nPr · NO → nCr. Everything else follows from this one decision.

Braces { } = set = nCr. Round brackets ( ) = ordered list = nPr.

The nPr → nCr conversion

Used nPr but then noticed the positions are interchangeable? Just divide by r!.

This one move converts any permutation answer into a combination answer. Very common in exams.

Pigeonhole — the recipe

h(k−1) + 1, where h = number of categories, k = how many needed in one.

If no specific k is mentioned, k is almost always 2, giving the simple h + 1.

Trigger word: "guarantee", "must", "at least … share", "find the minimum number so that".

"No two together" — the gap method

1. Arrange the other group: n!
2. m items in a row create m + 1 gaps
3. Put one of each into distinct gaps: P(m+1, k)
4. Multiply

Round table: answer is (n−1)! because rotating the whole arrangement gives the same one. Fix one person, then arrange the rest. Add another division by 2 if mirror images also count as the same.

Choosing r from n — use the small index

ⁿC r = ⁿC ⁿ⁻ʳ. Always use whichever index is smaller — it is less arithmetic and less chance of an error. This is why ⁷C 2 = ⁷C 5 = 21.

Trigger word → formula table

Pascal's triangle

Row n of the triangle IS the list nC 0, nC 1, …, nC n. Nothing more to it.

In (a+b)ⁿ, the exponent of b gives you r. So ⁷C 2, not ⁷C 5 — but they are equal, so pick the small one.

Check: put a = b = 1. The sum of the row must equal 2ⁿ.

THE master rule

ADD for "or". MULTIPLY for "and".

Overlapping events → subtract the overlap when adding. Dependent events → use the conditional when multiplying.

The complement trick

P(Ā) = 1 − P(A). Fire this on every question containing: "at least one", "at least a", "not", "no", "none", "more than", "fewer than", "everything except".

Example: "at least one 6 in 3 throws" → 1 − (5/6)³. Doing it directly would need 4 separate cases.

Independent or dependent? Look for one word

• "with replacement", "replace the card", "reset" → INDEPENDENT
• "without replacement", "no replacement", "used", "removed" → DEPENDENT

That single word decides which formula you use. Train yourself to spot it instantly.

Three questions, always in this order

1. Can both happen together? (decides whether to subtract an overlap)
2. Does the first change the second? (decides P(B) vs P(B|A))
3. Is a condition stated? "given that", "if we know", "suppose" (decides the denominator)

Conditional probability — shrink the space

Write out the new sample space in full, cross out everything the condition forbids, then count. Slower, but almost never wrong, and it earns the method mark.

Sanity check: the conditional probability must be between 0 and 1 and usually larger than the unconditional one, because you have removed the impossible cases.

Bayes in three lines — "target, split, divide"

1. Target on top: write P(A|B) at the top of your working.
2. Split below: P(B) = P(B|A)P(A) + P(B|Ā)P(Ā). Never leave a bare P(B).
3. Divide last.

Direction trap: P(A|B) ≠ P(B|A) unless P(A) = P(B). This is the classic exam trap. Read the question twice before you start.

p-hat vs p-nought

p̂ (p-hat) = the value you observed (it is an estimate of the truth).

p₀ (p-nought) = the value in H₀ (what you are nillying out).

In the standard error you always use p₀, never p̂.

Breaking a big binomial into pieces

"4 boys and 3 girls, pick 3 with at least one girl" — split by exactly how many girls, then add:

The case-by-case method also answers "exactly one" questions, which subtraction cannot.

Bars vs areas

Binomial = bars (P(X=3) is a bar's height). Normal = areas (P(X<3) is an area).

Consequence: P(X = c) = 0 for a normal distribution. A single point has zero width. You can never ask for "exactly" on a continuous variable — always a range.

68 – 95 – 99.7

Within 1σ, 2σ, 3σ of the mean. Memorise this and you can skip the z-table entirely for "how much data lies within k standard deviations" questions.

Start from 0.5, never from 0

The z-table only stores Φ(z) = the area from 0 to z. Every answer is built from that plus or minus 0.5:

• P(X < z) = Φ(z) + 0.5
• P(0 < X < z) = Φ(z) − 0.5
• P(X > z) = 0.5 − Φ(z)

Also useful: the curve is symmetric, so Φ(−z) = 1 − Φ(z). Look for a negative z and flip it to positive before going to the table.

The variance trap

N(μ, σ²) means the second number is the variance. If you are given 25, then σ = √25 = 5, not 25.

Rule: if the second number is a perfect square, it is almost certainly a variance. If it is not a square, it is probably σ directly.

Binomial checklist — one "no" kills the question

1. Fixed number of trials n
2. Only two outcomes (success/failure)
3. Independent trials
4. Constant probability p

"Without replacement" quietly breaks #3 and #4. That is the usual reason a question is not binomial.

Remember the shape of the binomial, not just the formula

• p near ½ → balanced, tall, biggest σ (since p(1−p) peaks at p = 0.5)
• p near 0 or 1 → lopsided, flat, hugging a corner

If your drawn bar chart disagrees with p, you have a mistake.

Continuity correction — "shave half off the outside"

Every boundary moves 0.5 outward, because a bar has width and you need to cover the whole bar, not the gap between bars.

The 5-step approximation routine

1. Check np ≥ 5 and n(1−p) ≥ 5 — write this down, it is a mark
2. Write μ = np, σ = √np(1−p) and keep them visible
3. Correct the boundary by 0.5, writing the corrected version in full
4. Standardise to z
5. Read the table, adding or subtracting 0.5

Sanity checks

• Answer between 0 and 1 — always.
• "At least the mean" should give something a bit above 0.5.
• A "less than mean" question should give something below 0.5.
• Symmetric question (μ ± kσ) should give roughly the 68–95–99.7 figure.

"Two tails ⇒ 2"

At α = 0.05: one-tailed is 1.645, two-tailed is 1.96.

The two-tailed value is always the bigger number. If you cannot remember which is which, remember that. Writing the wrong one is the single biggest error in this unit.

Choose the tail from the verb

• "increased", "more than", "greater", "rose" → right-tailed, one-tail, H₁: μ > μ₀
• "decreased", "less than", "fell", "reduced" → left-tailed, one-tail, H₁: μ < μ₀
• "changed", "different", "differs", "is it correct" → two-tail, H₁: μ ≠ μ₀

No direction mentioned → two-tailed. When in doubt, two-tailed is the safe default.

z or t?

• σ known, or n ≥ 30 → z-test
• σ unknown, or n < 30 → t-test, df = n − 1

The t critical value is always larger. If you used t but the question gave σ, you will usually still get the same decision — so a t-test is a safe guess if you are unsure. But state which one you used and why.

One shape, two tests

Both are (observed − claimed) ÷ standard error. Write it that way in your notes and you will never mix up the numerator or the denominator.

The 6-step recipe — this is the marking scheme

1. State H₀ and H₁ [2]
2. Name α and the tail type [1]
3. Critical value and critical region [1]
4. Test statistic, showing the SE separately [2]
5. Decision: REJECT or FAIL TO REJECT [1]
6. Conclusion in context, quoting α [1]

Region vs Value — the words are not interchangeable

Region = a range ("z > 1.645", or the shaded area). Value = one number (1.645).

"Find the critical values" wants numbers. "State the critical region" wants a range. This swap costs marks every year.

"Fail to reject", never "accept"

Not finding evidence against H₀ does not prove H₀. Write "there is insufficient evidence to reject H₀". This is a standard marking point.

Courtroom memory hook for errors

• Type I = convicting the innocent. Rejecting a true H₀. Probability = α. You took action against someone who was actually fine.
• Type II = letting the guilty go. Failing to reject a false H₀. Probability = β. You missed something real.

Trade-off: lower α ⇒ fewer Type I but more Type II. You cannot have both small.

The 3-line pre-submit check

1. Did I halve α for a two-tailed test? (1.96, not 1.645?)
2. Did I use p₀ and σ in the denominator, not p̂ or x̄?
3. Did I write "fail to reject" and never "accept"?

These three catch the overwhelming majority of lost marks.

The parity check — 10 seconds, free mark

Add up all the degrees first. If the total is even, plausible — edges = total/2. If odd, no such graph exists. Say so and stop.

Corollary: there is always an even number of odd-degree vertices.

Length counts EDGES, not vertices

A–B–C is length 2, even though it visits three vertices. Count the lines.

n − 1 is the number to remember

• Connected graph with n vertices needs at least n − 1 edges
• More than n − 1 → it has a cycle
• A tree has exactly n − 1 edges
• A spanning tree has exactly n − 1 edges
• Edges to delete to make a tree: m − n + 1

Adjacent vs Incident

Adjacent = vertex ↔ vertex (two dots are neighbours).

Incide/Incidence = vertex ↔ edge (a dot is on a line).

So: Adjacency matrix = n × n (all points). Incidence matrix = n × m (points × lines).

Matrix sanity checks

• Undirected ⇒ adjacency matrix is symmetric
• Row sum = degree (undirected) or out-degree (directed)
• Incidence matrix: every column sums to 2 in an undirected graph
• Diagonal = 0 (no loops), 2 if there is a loop

If a matrix you built fails one of these, you made an error — this catches mistakes without needing the original graph.

Dijkstra on paper

Use a table with permanent (ink) and tentative (pencil) columns.

1. Pick the smallest tentative label
2. Move it to permanent — it never changes again
3. Tentative = min(itself, permanent + edge weight)
4. Stop when all n vertices are permanent

Dijkstra needs non-negative weights. Negative edges → use Bellman–Ford.

Dijkstra = cheapest route (weighted). BFS = fewest hops (unweighted). No numbers on the edges? You want BFS.

Colouring in one test

Try to 2-colour the graph. Success ⇒ bipartite, χ = 2. Failure ⇒ there's an odd cycle somewhere, so χ ≥ 3.

Fast odd-cycle hunt: trace any closed loop. An odd number of edges (3, 5, 7…) means not bipartite. This catches most exam items in seconds.

Memorise: even cycle = 2, odd cycle = 3, Kₙ = n, planar ≤ 4, tree = 2.

Traversal mnemonic

PRE = you are PRE-sented, you greet first → root first

IN = you meet them IN the middle → root in the middle

POST = introductions come POST-poned → root last

PRE = Node, Left, Right · IN = Left, Node, Right · POST = Left, Right, Node

Three 2-second checks on any traversal

• Pre-order starts with the root. Post-order ends with the root.
• All three contain exactly the same n values, just reordered.
• In-order on a BST is ascending. If yours is not, you erred.

The BST shortcut

Given a binary search tree and asked for in-order: skip the traversal and just write the values in ascending order. Free marks, and it cannot go wrong.

Traversal habits that prevent errors

• Always go left first. Keep going left until you hit nothing, then record.
• Cross out each node as you record it — this is the surest way to avoid recording one twice.
• Pre- and post-order work on any tree shape. In-order needs a binary tree — if in-order is asked, the tree is binary.

Kruskal vs Prim — the one-liner

Kruskal = Kheapest edges from anywhere. Prim = Pick one vertex and gProw from it.

Does the question name a starting vertex? → Prim. Otherwise Kruskal is safe. If you are unsure, either is accepted — pick the one you can do cleanly.

Both stop at exactly n − 1 edges, and on a weighted graph they give the same total. That is a free cross-check on your arithmetic.

Kruskal's cycle test

Ask: "are these two ends already in the same group?" If yes, adding the edge creates a cycle — skip it. Draw the groups as little circles; it is the union–find procedure and it cannot fail.

Write the sorted edge list down before you start — the examiner is looking for it.

Prim's edge test

At each step only edges with exactly one end inside the current tree are eligible. Edges with both ends inside are already used up (or would cycle). Edges with neither end inside are not reachable yet.

Complete tree sizes

Memorise 1, 3, 7, 15, 31, 63 for heights 0–5. Then min height for n nodes is ⌊log₂ n⌋ in three seconds.

The 5-step approach to any question

1. Classify it. Counting → Unit 2. Chance → Unit 3. Shape of data → Unit 4. "Significant?" → Unit 5. Network → Unit 6/7. Vocabulary → Unit 1.
2. Hunt the trigger word. "at least one", "no two", "guarantee", "cheapest", "different", "changed", "without replacement". Each one flips the method.
3. Write the formula BEFORE the numbers. Naming the formula is often a mark on its own.
4. Show one line of justification. In this syllabus, marks are for reasons far more often than for answers.
5. Sanity check. Probability in [0,1]? Counts whole? Degrees even? Tail count right?

Answers that earn method marks

• Write the formula symbolically, then substitute.
• Say why the events are independent (or not).
• Justify that the graph is a tree (connected + n − 1 edges).
• State that the binomial conditions are satisfied — or that they are not.
• For MST: state the number of edges at the end (n − 1).
• Interpret a negative exponent, a probability, or a degree in words once.