Skip to main content
Sign in Create a free account
Varsity Tutors
Sign in
Back to lessons

IB Mathematics: Analysis and Approaches • Number and Algebra

Combinations, Permutations & Extended Binomial — AHL 1.10 Combinations and permutations; extension of binomial theorem (n ∈ ℚ)

Master counting techniques and extend the binomial theorem beyond positive integers to rational exponents.

Explore the lesson 14 min read
IB Mathematics Analysis and Approaches emblem with a function, curved graph, and sigma notation
Inside this lesson 10 sections 5 questions 8 figures

SECTION 1

Historical Context & Motivation

Counting the number of ways to arrange or select objects might seem like a simple task, but it has driven some of the deepest discoveries in mathematics. Ancient civilizations wrestled with problems of arrangement, and their solutions eventually crystallised into the formal theories of permutations and combinations that you will study in this lesson. These counting techniques form the backbone of probability, statistics, and the binomial theorem — a formula that connects algebra with combinatorics in a surprisingly elegant way.

c. 200 BCE • Indian Combinatorics

The Indian mathematician Pingala studied arrangements of syllables in Sanskrit poetry. His work introduced an early form of Pascal's triangle and binary-like counting methods, laying a foundation for combinatorial reasoning.

1070 • Omar Khayyám's Triangle

Persian mathematician Omar Khayyám described the arithmetic triangle of binomial coefficients centuries before Pascal. He used it to find roots of polynomials and expand powers of sums.

1654 • Pascal's Traité

Blaise Pascal published his Traité du triangle arithmétique, formalising the triangle that now bears his name. His correspondence with Fermat on gambling problems also launched the modern theory of probability.

1665 • Newton's Generalisation

Isaac Newton extended the binomial theorem to fractional and negative exponents. This breakthrough turned a finite sum into an infinite series and opened doors to calculus and analysis.

1713 • Bernoulli's Ars Conjectandi

Jacob Bernoulli unified permutations, combinations, and the binomial theorem within the framework of probability theory. His posthumous work became a cornerstone of mathematical statistics.

The central question that drives this topic is deceptively simple: In how many ways can we arrange or choose objects, and how do these counts show up when we expand expressions like (1 + x)ⁿ? When n is a positive integer, the binomial expansion is a finite sum and every coefficient is a familiar combination number. But Newton asked a bolder question — what happens when n is a fraction like ½ or −1? This extension is what elevates the binomial theorem from a handy algebraic shortcut to a powerful tool in analysis.

SECTION 2

Core Principles & Definitions

Before diving into formulas, you need a clear understanding of the foundational ideas. Permutations and combinations both count selections from a set, but they differ in one crucial way: whether the order of selection matters. The binomial theorem then connects these counting ideas directly to algebraic expansion.

1

Factorial (n!)

The product of all positive integers from 1 to n. For example, 5! = 5 × 4 × 3 × 2 × 1 = 120. By convention, 0! = 1. Factorials measure the total number of ways to arrange n distinct objects.
2

Permutations P(n, r)

The number of ordered arrangements of r objects chosen from n distinct objects. Order matters: choosing A then B is different from B then A. Formula: P(n, r) = n! / (n − r)!.
3

Combinations C(n, r)

The number of unordered selections of r objects from n. Order does not matter: {A, B} is the same as {B, A}. Formula: C(n, r) = n! / [r!(n − r)!]. Also written ⁿCᵣ or (n choose r).
4

Binomial Theorem (n ∈ ℤ⁺)

For positive integer n: (a + b)ⁿ = Σ from r = 0 to n of C(n, r) · aⁿ⁻ʳ · bʳ. This is a finite sum with exactly n + 1 terms. Each coefficient is a combination number from Pascal's triangle.
5

Extended Binomial (n ∈ ℚ)

Newton's extension: when n is rational (e.g., ½, −3), (1 + x)ⁿ becomes an infinite series valid for |x| < 1. The 'coefficients' are generalised using the formula n(n−1)(n−2)…(n−r+1) / r!.
✦

KEY TAKEAWAY

Think of permutations as arranging books on a shelf — the order you place them matters. Combinations are like picking a team from a group — once the team is chosen, who was picked first doesn't matter. The binomial theorem tells you how many times each combination appears when you expand a power. And Newton's extension is like discovering that this counting logic works even when the 'power' is a fraction, producing an infinite series instead of a finite one.

SECTION 3

Visual Explanation — Pascal's Triangle & Counting

Pascal's triangle is one of the most beautiful structures in mathematics. Each entry is a combination number C(n, r), and it also gives you the coefficients when you expand (a + b)ⁿ for positive integer n. The diagram below shows the first six rows and highlights the connection between combinations, permutations, and the binomial expansion.

Pascal's Triangle — Binomial Coefficients C(n, r)11112113311464115101051n = 0n = 1n = 2n = 3n = 4n = 5Reading the TriangleRow n = 4:(a+b)⁴ = 1a⁴ + 4a³b + 6a²b² + 4ab³ + 1b⁴Coefficients: C(4,0)=1, C(4,1)=4, C(4,2)=6, C(4,3)=4, C(4,4)=1Each entry = sum of the two entries directly above it.
Pascal's triangle displays binomial coefficients C(n, r). Each row corresponds to a power n, and each entry gives the coefficient of the corresponding term in (a + b)ⁿ. Notice that each number is the sum of the two numbers directly above it.

Notice how the entries in row n = 4 — which are 1, 4, 6, 4, 1 — appear directly as the coefficients in the expansion of (a + b)⁴. This is no coincidence. The coefficient of the term aⁿ⁻ʳbʳ is precisely C(n, r), the number of ways to choose which r factors (out of n brackets) contribute a b instead of an a. This combinatorial interpretation is the key insight connecting counting and algebra.

SECTION 4

Mathematical Framework

Now let's formalise the key formulas. We begin with the counting formulas for permutations and combinations, then state the standard binomial theorem, and finally present Newton's extension to rational exponents.

PERMUTATIONS

P(n,r)=n!/(n−r)!P(n, r) = n! / (n − r)!

n = total number of distinct objects; r = number of objects selected. This counts ordered arrangements. Example: P(5, 3) = 5! / 2! = 120 / 2 = 60.

COMBINATIONS

C(n,r)=n!/[r!×(n−r)!]C(n, r) = n! / [r! × (n − r)!]

Also written as ⁿCᵣ or (n choose r). This counts unordered selections. We divide the permutation count by r! to remove the duplicate orderings. Example: C(5, 3) = 60 / 6 = 10.

BINOMIAL THEOREM (n ∈ ℤ⁺)

(a+b)n=Σ(r=0ton)C(n,r)⋅an−r⋅br(a + b)ⁿ = Σ (r = 0 to n) C(n, r) · aⁿ⁻ʳ · bʳ

For positive integer n, this is a finite sum with n + 1 terms. Each term's coefficient is a combination number from Pascal's triangle.

EXTENDED BINOMIAL THEOREM (n ∈ ℚ)

(1+x)n=1+nx+[n(n−1)/2!]x2+[n(n−1)(n−2)/3!]x3+…(1 + x)ⁿ = 1 + nx + [n(n−1)/2!]x² + [n(n−1)(n−2)/3!]x³ + …

Valid for |x| < 1 when n ∈ ℚ. The generalised binomial coefficient is defined as: C(n, r) = n(n−1)(n−2)…(n−r+1) / r!. When n is not a positive integer, the series is infinite and converges only when |x| < 1.

⚠

Convergence Condition

The condition |x| < 1 is essential for the extended binomial theorem. Without it, the infinite series does not converge to a finite value. In IB exams, always state this condition when using the extended form. Also note that the expression must be in the form (1 + x)ⁿ — you may need to factorise first if it starts as (a + b)ⁿ.

SECTION 5

The Extended Binomial Theorem in Detail

The standard binomial theorem works perfectly when n is a positive integer — you get a neat polynomial with n + 1 terms. But what happens when n = ½, or n = −1? Newton's genius was to show that the same pattern of coefficients still works, except now the series never terminates. Let's see this in action by examining how the generalised binomial coefficient differs from the standard one.

Standard vs Extended Binomial TheoremStandard (n ∈ ℤ⁺)n = 3:(1+x)³ = 1 + 3x + 3x² + x³✓ Finite: 4 terms✓ Valid for all x✓ Coefficients from Pascal's triangleC(3,r) = 3!/(r!(3−r)!)Extended (n ∈ ℚ)n = ½:(1+x)^½ = 1 + ½x − ⅛x² + ¹⁄₁₆x³ − ...✗ Infinite series (never terminates)✓ Valid only for |x| < 1✓ Generalised coefficientsC(½,r) = ½(½−1)(½−2)…/r!Generalised Coefficient Computation (n = −1, r = 3)C(−1, 3) = (−1)(−1−1)(−1−2) / 3! = (−1)(−2)(−3) / 6 = −6/6 = −1C(½, 2) = (½)(½−1) / 2! = (½)(−½) / 2 = −¼ / 2 = −⅛C(−2, 3) = (−2)(−3)(−4) / 3! = −24 / 6 = −4
The left panel shows the standard binomial expansion (n = 3), which terminates after 4 terms and is valid for all x. The right panel shows the extended version (n = ½), which produces an infinite series valid only for |x| < 1. Below, three examples show how to compute generalised binomial coefficients when n is rational or negative.

A critical detail is how the generalised coefficient is computed. For a positive integer n, the product n(n−1)(n−2)…(n−r+1) eventually reaches 1 and stops, giving the familiar factorial ratio. But for n = ½, the factors become ½, −½, −³⁄₂, −⁵⁄₂, and so on — they never reach zero, so the series never terminates. The signs alternate in many cases (as with n = ½), which is something you should watch carefully in exam calculations.

💡

IB Exam Tip

When expanding (1 + x)ⁿ for rational n, most IB questions ask for the first three or four terms. Write each generalised coefficient as a fraction, simplify carefully, and always state the validity condition |x| < 1. If the expression is (a + bx)ⁿ, first factor out aⁿ to obtain aⁿ(1 + (b/a)x)ⁿ, then apply the formula to (1 + (b/a)x)ⁿ with the condition |bx/a| < 1.

SECTION 6

Worked Examples

Example 1: Permutations & Combinations

A committee of 4 is chosen from 10 students. (a) In how many ways can a President, Vice-President, Secretary, and Treasurer be assigned? (b) In how many ways can a committee of 4 be chosen with no assigned roles?

  1. 1

    Step 1 — Identify whether order matters

    Part (a) involves assigning distinct roles, so order matters → use permutations. Part (b) involves choosing a group with no distinction among members, so order does not matter → use combinations.
  2. 2

    Step 2 — Calculate part (a) using P(n, r)

    P(10, 4) = 10! / (10 − 4)! = 10! / 6! = 10 × 9 × 8 × 7 = 5040.
    P(10, 4) = 5040 ways
  3. 3

    Step 3 — Calculate part (b) using C(n, r)

    C(10, 4) = 10! / [4! × 6!] = 5040 / 24 = 210. Notice that we divide the permutation count by 4! = 24 because each group of 4 people can be arranged in 24 ways, all of which represent the same committee.
    C(10, 4) = 210 ways

Example 2: Extended Binomial Theorem

Find the first four terms of (1+x)−2(1 + x)^{−2} and state the values of x for which the expansion is valid.

  1. 1

    Step 1 — Identify n and apply the generalised formula

    Here n = −2. Since n is not a positive integer, we use the extended binomial theorem: (1 + x)ⁿ = 1 + nx + [n(n−1)/2!]x² + [n(n−1)(n−2)/3!]x³ + …
  2. 2

    Step 2 — Compute the generalised coefficients

    First term: 1. Second term coefficient: n = −2. Third term: n(n−1)/2! = (−2)(−3)/2 = 6/2 = 3. Fourth term: n(n−1)(n−2)/3! = (−2)(−3)(−4)/6 = −24/6 = −4.
  3. 3

    Step 3 — Write out the expansion

    (1 + x)⁻² = 1 + (−2)x + 3x² + (−4)x³ + … = 1 − 2x + 3x² − 4x³ + …
    (1 + x)⁻² = 1 − 2x + 3x² − 4x³ + …
  4. 4

    Step 4 — State the validity condition

    The series converges when |x| < 1. This is essential — if |x| ≥ 1, the terms grow without bound and the series diverges.
    Valid for |x| < 1

Example 3: Extended Binomial with Factoring

Find the first three terms of (4 + x)^(½).

  1. 1

    Step 1 — Factor to get the form (1 + u)ⁿ

    (4 + x)^(½) = [4(1 + x/4)]^(½) = 4^(½) × (1 + x/4)^(½) = 2(1 + x/4)^(½). Now let u = x/4 and n = ½.
  2. 2

    Step 2 — Expand (1 + u)^(½) to three terms

    (1 + u)^(½) ≈ 1 + ½u + [½(½−1)/2!]u² = 1 + ½u + [(½)(−½)/2]u² = 1 + ½u − ⅛u².
  3. 3

    Step 3 — Substitute back u = x/4 and multiply by 2

    2[1 + ½(x/4) − ⅛(x/4)²] = 2[1 + x/8 − x²/128] = 2 + x/4 − x²/64.
    (4 + x)^(½) ≈ 2 + x/4 − x²/64
  4. 4

    Step 4 — Validity condition

    |x/4| < 1 ⟹ |x| < 4. The expansion is valid for |x| < 4.
    Valid for |x| < 4

SECTION 7

Permutations vs Combinations — When to Use Which

One of the most common mistakes in counting problems is confusing permutations with combinations. The table below provides a clear framework for deciding which formula to use. The key question is always: does the order of selection affect the outcome?

FeaturePermutations P(n, r)Combinations C(n, r)
OrderMatters (AB ≠ BA)Does not matter (AB = BA)
Formulan! / (n − r)!n! / [r!(n − r)!]
Result sizeLarger (more arrangements)Smaller (groups not orderings)
RelationshipP(n, r) = r! × C(n, r)C(n, r) = P(n, r) / r!
Typical scenariosRankings, PIN codes, seating arrangements, race finishesCommittees, teams, card hands, lottery selections

Comparison of permutations and combinations

✦

KEY TAKEAWAY

Imagine you're at a pizza shop choosing 3 toppings from 8 options. You don't care what order you pick them — pepperoni, mushrooms, onions is the same pizza as onions, pepperoni, mushrooms. That's a combination. Now imagine you're setting a 3-digit lock code from digits 1–8 (no repeats). Here 1-3-5 is a completely different code from 5-3-1. That's a permutation. Same set of numbers, different counting methods — all because of whether arrangement matters.

SECTION 8

Connection to Advanced Theory

The extended binomial theorem is not just an algebraic curiosity — it bridges directly into topics you will encounter in further mathematics and university-level analysis. The table below shows how the ideas in AHL 1.10 connect to more advanced concepts.

AHL 1.10 ConceptAdvanced Extension
Combinations C(n, r)Multinomial coefficients for expanding (a + b + c + …)ⁿ; combinatorial identities used in discrete mathematics and probability.
Binomial theorem (n ∈ ℤ⁺)Binomial distribution in statistics: P(X = r) = C(n, r)pʳ(1−p)ⁿ⁻ʳ; connects directly to probability theory.
Extended binomial (n ∈ ℚ)Taylor and Maclaurin series expansions; power series representations of functions like √(1+x), 1/(1+x)², (1+x)^(−½).
Convergence condition |x| < 1Radius of convergence for power series; ratio test and root test in infinite series analysis.
Generalised binomial coefficientsGamma function Γ(n+1) = n! generalised to all complex numbers; used in advanced probability distributions.

How AHL 1.10 concepts extend into higher mathematics

One of the most important applications within the IB curriculum itself is the connection between the binomial theorem and the binomial distribution in the Statistics and Probability topic. The coefficients C(n, r) that you compute for algebraic expansion are the same coefficients that determine the probability of getting exactly r successes in n independent trials. Meanwhile, the extended binomial theorem connects to Maclaurin series (Topic 5), where you will learn to approximate functions using infinite polynomials — a technique built on exactly the same generalised coefficients you study here.

ℹ

Looking Ahead

If you continue to study mathematics at university, you will find that the extended binomial theorem is a special case of the Taylor series expansion about x = 0. The condition |x| < 1 is a specific instance of the radius of convergence — a concept that unifies all power series. Mastering AHL 1.10 gives you a genuine head start on these ideas.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain, in your own words, why C(n, r) is always less than or equal to P(n, r) for the same values of n and r. Under what condition are they equal?
Show answer
C(n, r) counts unordered groups while P(n, r) counts ordered arrangements of the same selections. Since every unordered group of r objects can be arranged in r! different orders, we have P(n, r) = r! × C(n, r). This means C(n, r) = P(n, r) / r!, so C(n, r) ≤ P(n, r). They are equal only when r = 0 or r = 1, because 0! = 1 and 1! = 1, meaning dividing by r! doesn't change the value.

Question 2: BASIC CALCULATION

A class has 12 students. (a) How many ways can a president, vice-president, and secretary be chosen? (b) How many ways can a committee of 3 be chosen?
Show answer
(a) Since roles are assigned, order matters: P(12, 3) = 12 × 11 × 10 = 1320 ways. (b) Since the committee has no assigned roles, order does not matter: C(12, 3) = 12! / (3! × 9!) = 1320 / 6 = 220 ways. Notice that dividing the permutation count by 3! = 6 eliminates the duplicate orderings of each group of 3 people.

Question 3: INTERMEDIATE

Find the first four terms of (1−2x)−3(1 − 2x)^{−3} using the extended binomial theorem. State the values of x for which the expansion is valid.
Show answer
Write (1−2x)−3(1 − 2x)^{−3} = (1+(−2x))−3(1 + (−2x))^{−3}, so n = −3 and the substitution is u = −2x. The first four terms are: 1 + (−3)(−2x) + [(−3)(−4)/2!](−2x)² + [(−3)(−4)(−5)/3!](−2x)³ = 1 + 6x + [(12/2)(4x²)] + [(−60/6)(−8x³)] = 1 + 6x + 24x² + 80x³. The validity condition requires |−2x| < 1, which gives |x| < ½.

Question 4: APPLIED

A basketball coach needs to choose 5 starting players from a squad of 14. She has decided that one specific player (the captain) must always start. How many different starting line-ups are possible?
Show answer
Since the captain is always included, we need to choose the remaining 4 players from the other 13 squad members. The number of ways is C(13, 4) = 13! / (4! × 9!) = (13 × 12 × 11 × 10) / (4 × 3 × 2 × 1) = 17160 / 24 = 715. Note that we use combinations because the 5 starting positions are not ranked — the question only asks which players are on the court, not their specific roles.

Question 5: CRITICAL THINKING

Use the extended binomial theorem to show that √(1.02) ≈ 1.01 − 0.00005 (to five decimal places) by expanding (1 + x)^(½) with x = 0.02. Find the first three terms and discuss why this approximation is so accurate.
Show answer
Set x = 0.02 in (1 + x)^(½) = 1 + ½x + [(½)(−½)/2!]x² + … The first three terms give: 1 + ½(0.02) + (−⅛)(0.02)² = 1 + 0.01 + (−0.125)(0.0004) = 1 + 0.01 − 0.00005 = 1.00995. A calculator gives √(1.02) = 1.009950494…, confirming the result is accurate to five decimal places. The approximation works so well because x = 0.02 is much smaller than 1 (satisfying |x| < 1 easily), which means higher-order terms like x³ = 0.000008 are negligibly small. Each successive term shrinks rapidly, so just two or three terms capture almost all of the true value.

SUMMARY

Lesson Summary

Summary

This lesson covered three interconnected ideas from AHL 1.10. Permutations count ordered arrangements using P(n, r) = n!/(n−r)!, while combinations count unordered selections using C(n, r) = n!/[r!(n−r)!]. The relationship P(n, r) = r! × C(n, r) shows that permutations always give a count that is r! times larger than combinations, since each unordered group can be arranged in r! different ways. These combination numbers appear as coefficients in the binomial theorem for positive integer exponents: (a + b)ⁿ = Σ C(n, r) · aⁿ⁻ʳ · bʳ, which is a finite sum with n + 1 terms.

Newton's extended binomial theorem generalises this to rational exponents n ∈ ℚ: (1 + x)ⁿ = 1 + nx + [n(n−1)/2!]x² + [n(n−1)(n−2)/3!]x³ + …, producing an infinite series that converges only when |x| < 1. The generalised binomial coefficient n(n−1)(n−2)…(n−r+1)/r! replaces the standard C(n, r). Remember to factorise expressions into the form aⁿ(1 + u)ⁿ before applying the formula, and always state the validity condition in your answers. These concepts connect to the binomial distribution in probability and to Maclaurin series in calculus, making them foundational tools throughout the IB Mathematics course.

Varsity Tutors • IB Mathematics: Analysis and Approaches • Combinations, Permutations & Extended Binomial