This HL mini-lesson extends Number & Algebra: counting principles, permutations and combinations, the binomial theorem and its general term, partial fractions, a first look at complex numbers, and proof by mathematical induction.
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Counting
The counting principles
If a first task can be done in m ways and a second in n ways, together they can be done in m × n ways (product rule). Distinct arrangements of all n objects number n!.
n! = n(n−1)(n−2)…2·1with 0! = 1 by definition
Quick check
Order or not?
?A club elects a president, then a vice-president, then a secretary from 7 members. Counting the possible outcomes is a:
Permutations
Permutations
A permutation counts ordered selections. The number of ways to arrange r objects from n is:
ₙPᵣ = n! / (n − r)!e.g. ₇P₃ = 7·6·5 = 210
Order matters — "president, then vice-president" is different from the reverse.
Calculate
Permutations
✎How many ways can 3 people be arranged from 7 (i.e. ₇P₃)?
Hint: ₇P₃ = 7 × 6 × 5.
Combinations
Combinations
A combination counts unordered selections:
ₙCᵣ = n! / (r!(n − r)!) = ₙPᵣ / r!e.g. ₈C₃ = 56 — a committee, not a ranking
ₙCᵣ are exactly the binomial coefficients in row n of Pascal's triangle.
Quick check
Imaginary unit
?In the complex numbers, i² equals:
Calculate
Combinations
✎Evaluate ₈C₃ (choosing 3 from 8, order not important).
Hint: ₈C₃ = 8!/(3!·5!) = (8×7×6)/(3×2×1).
Binomial
The binomial theorem & general term
For positive integer n:
(a+b)ⁿ = Σr=0n ₙCᵣ aⁿ⁻ʳ bʳ
General term
The term in bʳ is ₙCᵣ aⁿ⁻ʳ bʳ. In (1 + 2x)⁶ the x³ term is ₆C₃(2x)³ = 20·8 x³ = 160x³.
Sort it
Permutation, combination or neither?
Tap a scenario, then how you would count it.
🔢 Permutation
🎯 Combination
➖ Neither
Partial fractions
Partial fractions
A proper rational function with distinct linear factors splits into simpler pieces — vital for later integration:
(3x+5)/((x+1)(x+2)) = A/(x+1) + B/(x+2)cover-up: x = −1 gives A = 2; x = −2 gives B = 1
Calculate
Partial fractions
✎For (3x+5)/((x+1)(x+2)) = A/(x+1) + B/(x+2), find A.
Hint: multiply out and set x = −1 to isolate A: (3(−1)+5)/(−1+2).
Quick check
Induction step
?After proving the base case, the inductive step requires you to:
Complex numbers
A first look at complex numbers
The imaginary unit satisfies i² = −1. A complex number z = a + bi has real part a and imaginary part b.
i² = −1add/subtract componentwise; multiply using i² = −1
Conjugate: z* = a − bi; z·z* = a² + b² is real. (Explored fully in the Complex Numbers lesson.)
Calculate
Binomial term
✎Find the coefficient of x³ in the expansion of (1 + 2x)⁶.
Hint: ₆C₃ (2x)³ = 20 × 8 × x³.
Match it
Match each expression to its value
Tap a statement on the left, then its matching partner on the right.
Expression
Value
Induction
Proof by mathematical induction
To prove a statement P(n) for all integers n ≥ 1:
Base case: show P(1) is true.
Inductive step: assume P(k) true, then prove P(k+1) true.
Example fact: Σr=1n r = n(n+1)/2, so at n = 100 the sum is 100·101/2 = 5050.
Calculate
Induction sum
✎Using Σ r = n(n+1)/2, find the sum of the first 100 positive integers.
Hint: 100 × 101 ÷ 2.
Quick check
Partial fractions
?Partial fractions with distinct linear factors can be applied to:
Strategy
Counting & proof strategy
Ask "does order matter?" — yes → permutation, no → combination. For a binomial coefficient, pick the r giving the required power. For induction, state the assumption and target clearly.
Exam habit: in induction always finish with a concluding sentence "true for n=k+1, hence by induction true for all n".
Recap
The big ideas to know
Counting: product rule; n! arrangements; 0! = 1
Permutations: ₙPᵣ = n!/(n−r)! — order matters
Combinations: ₙCᵣ = n!/(r!(n−r)!) — order does not
Binomial: general term ₙCᵣ aⁿ⁻ʳ bʳ; partial fractions split proper fractions
Induction: base case P(1) + inductive step P(k) ⇒ P(k+1)
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