Combinatorics Basics — SL 1.9 Combinatorics basics (nCr as used in binomial theorem) - IB Mathematics Analysis and Approaches

SECTION 1

Historical Context & Motivation

Long before modern algebra existed, mathematicians across the world were fascinated by a deceptively simple question: how many ways can you choose a smaller group from a larger one? Whether it was ancient Indian scholars counting poetic metres or Chinese mathematicians arranging objects, the art of combinatorics — the mathematics of counting — has roots stretching back thousands of years. The specific tool we focus on today, the binomial coefficient written as ⁿCᵣ, tells us exactly how many ways we can pick r items from a set of n items when order doesn't matter. This idea is central to the binomial theorem and appears throughout probability, statistics, and algebra.

~200 BCE • Pingala's Chandaḥśāstra

The Indian mathematician Pingala studied combinations of syllable patterns in Sanskrit poetry, producing one of the earliest known references to the numbers we now call binomial coefficients.

1070 • Omar Khayyám's Triangle

Persian mathematician Omar Khayyám described a triangular arrangement of binomial coefficients and used it to extract roots of numbers, centuries before European mathematicians independently discovered the same structure.

1303 • Zhu Shijie's Precious Mirror

Chinese mathematician Zhu Shijie published the triangular array of coefficients in his work Precious Mirror of the Four Elements, showing deep understanding of the patterns within combinations.

1654 • Pascal's Traité

Blaise Pascal published his Traité du triangle arithmétique, formalising the triangle that now bears his name and connecting it to probability theory through his famous correspondence with Fermat.

1665 • Newton's Generalisation

Isaac Newton extended the binomial theorem to non-integer and negative exponents, dramatically widening the reach of ⁿCᵣ and laying groundwork for calculus and infinite series.

The question these mathematicians all circled back to was the same one you will master in this lesson: given n objects, how many distinct groups of r can you form? And once you can answer that, how does it help you expand expressions like (a + b)ⁿ without multiplying everything out by hand?

SECTION 2

Core Principles & Definitions

Before diving into formulas, you need a solid grasp of the building blocks. Combinatorics rests on a few key ideas that are easy to state but powerful when combined. Each concept below feeds directly into the formula for ⁿCᵣ and, ultimately, into the binomial theorem.

Factorial (n!)

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

Permutations (ⁿPᵣ)

The number of ways to choose r items from n when order matters. Calculated as n!/(n − r)!. Choosing president, then vice-president from 10 people is a permutation problem.

Combinations (ⁿCᵣ)

The number of ways to choose r items from n when order does NOT matter. Calculated as n! / (r!(n − r)!). Choosing a committee of 3 from 10 people is a combination problem.

Binomial Coefficient Notation

ⁿCᵣ is also written as C(n, r) or with the 'n-choose-r' bracket notation. All three notations mean exactly the same thing: n! / (r!(n − r)!).

Symmetry Property

ⁿCᵣ = ⁿC₍ₙ₋ᵣ₎. Choosing which 3 people to include from 10 is the same as choosing which 7 to exclude. This symmetry is visible in every row of Pascal's Triangle.

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KEY TAKEAWAY

Think of combinations like choosing toppings for a pizza. If you pick pepperoni, mushrooms, and olives, the order you point at them doesn't create a different pizza — it's the same set of toppings. That's why combinations divide out the repeated arrangements by dividing the permutation count by r!. Permutations care about order; combinations do not.

SECTION 3

Pascal's Triangle — A Visual Map of ⁿCᵣ

One of the most elegant ways to see all the values of ⁿCᵣ at once is through Pascal's Triangle. Each row corresponds to a value of n, and each entry in that row is a specific ⁿCᵣ value. The triangle starts with ⁰C₀ = 1 at the top, and every interior number is the sum of the two numbers directly above it. This simple rule generates every binomial coefficient you will ever need.

Pascal's Triangle showing rows 0 through 6. Each circle contains the value of ⁿCᵣ, where the row number gives n and the position within the row (counting from 0) gives r. Notice the left-right symmetry in every row — this reflects the property ⁿCᵣ = ⁿC₍ₙ₋ᵣ₎.

Look at row 4 in the diagram: the entries are 1, 4, 6, 4, 1. These are exactly the coefficients you get when you expand (a + b)⁴ = 1a⁴ + 4a³b + 6a²b² + 4ab³ + 1b⁴. This is no coincidence — the entire purpose of Pascal's Triangle is to store the binomial coefficients that appear in the binomial theorem. The addition rule (each entry = the sum of the two above) also gives us a recursive way to compute ⁿCᵣ without using the factorial formula: ⁿCᵣ = ⁿ⁻¹Cᵣ₋₁ + ⁿ⁻¹Cᵣ.

SECTION 4

Mathematical Framework

Now let's formalise the ideas from the previous sections into precise formulas. You'll need these for IB exam questions, so make sure you can both recall them and understand why they work.

FACTORIAL

n! = n × (n − 1) × (n − 2) × … × 2 × 1

where n is a non-negative integer, and by definition 0! = 1. For example, 6! = 720.

PERMUTATIONS

ⁿPᵣ = n! / (n − r)!

Counts the number of ordered arrangements of r items chosen from n. This is the starting point before we 'remove' the effect of order.

COMBINATIONS (nCr)

ⁿCᵣ = n! / (r! × (n − r)!)

Also written C(n, r). This divides the permutation count by r! to remove duplicate orderings. It counts unordered selections of r items from n. This is the formula given in your IB formula booklet.

BINOMIAL THEOREM

(a + b)ⁿ = Σ (r = 0 to n) ⁿCᵣ × aⁿ⁻ʳ × bʳ

Each term in the expansion has the form ⁿCᵣ × aⁿ⁻ʳ × bʳ. The coefficients ⁿCᵣ tell you how many ways each combination of a's and b's arises. There are n + 1 terms in total, and r goes from 0 to n.

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Why divide by r! ?

Suppose you choose 3 students — Alice, Bob, and Cara — from a group of 10 to form a committee. The permutation formula counts ABC, ACB, BAC, BCA, CAB, CBA as six different selections. But it's the same committee! Those 6 rearrangements are 3! = 6, so we divide by 3! to collapse them into one. That's exactly what the r! in the denominator does.

SECTION 5

Connecting ⁿCᵣ to the Binomial Expansion

The real power of ⁿCᵣ in the IB syllabus is its role inside the binomial theorem. When you expand (a + b)ⁿ, each term involves picking some a's and some b's from n factors. The coefficient ⁿCᵣ counts how many ways to choose which r of the n factors contribute a b (the rest contribute an a). The following diagram shows this process for (a + b)⁴.

The expansion of (a + b)⁴ shown term by term. For each value of r, ⁴Cᵣ counts the number of ways to select which factors contribute b. The exponents on a and b always add up to 4.

Notice a few important patterns in the diagram. First, the exponents on a decrease from n to 0 while the exponents on b increase from 0 to n, and in every term the two exponents sum to n. Second, the coefficients are symmetric (1, 4, 6, 4, 1), matching the symmetry property ⁿCᵣ = ⁿC₍ₙ₋ᵣ₎ that we discussed earlier. Finally, the sum of all coefficients is 2ⁿ — in this case, 1 + 4 + 6 + 4 + 1 = 16 = 2⁴.

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Finding a Specific Term

A common IB question asks for a specific term. The general term (the (r + 1)th term) is Tᵣ₊₁ = ⁿCᵣ × aⁿ⁻ʳ × bʳ. Be careful: r starts from 0, so the 'third term' corresponds to r = 2.

SECTION 6

Worked Example

Let's walk through a complete example that combines ⁿCᵣ calculation with the binomial theorem.

Find the coefficient of x³ in the expansion of (2x + 3)⁵

1
Step 1 — Identify a, b, and n

In the expression (2x + 3)⁵, we set a = 2x, b = 3, and n = 5. We want the term that contains x³.
2
Step 2 — Determine the value of r

The general term is ⁵Cᵣ × (2x)⁵⁻ʳ × 3ʳ. The power of x in this term is 5 − r. We need 5 − r = 3, so r = 2.

r = 2
3
Step 3 — Calculate ⁵C₂

⁵C₂ = 5! / (2! × 3!) = (5 × 4) / (2 × 1) = 120 / (2 × 6) = 10.

⁵C₂ = 10
4
Step 4 — Compute the full term

The term is ⁵C₂ × (2x)³ × 3². Compute each piece: (2x)³ = 8x³ and 3² = 9. So the term = 10 × 8x³ × 9 = 10 × 72x³ = 720x³.

Term = 720x³
5
Step 5 — State the coefficient

The coefficient of x³ is 720.

Coefficient = 720

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STRATEGY SUMMARY

For any 'find the coefficient' problem: (1) identify a, b, n; (2) find the value of r that gives the correct power of the variable; (3) compute ⁿCᵣ; (4) multiply ⁿCᵣ by the appropriate powers of a and b. Don't forget that a and b may have their own coefficients (like the 2 in 2x), which must be raised to powers too.

SECTION 7

Permutations vs. Combinations — When to Use Which

One of the most common mistakes in combinatorics is confusing permutations with combinations. The table below clarifies when each applies and highlights the key differences.

Feature	Permutations (ⁿPᵣ)	Combinations (ⁿCᵣ)
Does order matter?	Yes — AB ≠ BA	No — {A, B} = {B, A}
Formula	n! / (n − r)!	n! / (r! × (n − r)!)
Relationship	ⁿPᵣ = ⁿCᵣ × r!	ⁿCᵣ = ⁿPᵣ / r!
Example scenario	Ranking top 3 out of 10 runners	Choosing a team of 3 from 10 players
Result (n=10, r=3)	720	120
Used in binomial theorem?	Not directly	Yes — as the coefficients

Permutations vs. Combinations comparison

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KEY TAKEAWAY

Here's a quick test: ask yourself, 'If I rearrange my selection, do I get something different?' If your answer is yes, use permutations. If no, use combinations. In the binomial theorem, the coefficients are always combinations because swapping which factors contribute an 'a' or 'b' doesn't change the resulting product — only the count matters.

SECTION 8

Connection to Advanced Topics

The ideas in SL 1.9 serve as a springboard into several more advanced areas of mathematics. Understanding ⁿCᵣ now will make these future topics far more accessible.

SL 1.9 Concept	Advanced Extension	Where You'll See It
ⁿCᵣ with positive integer n	Generalised binomial coefficients for any real n (Newton's extension)	HL Calculus, Taylor series
Pascal's Triangle addition rule	Pascal's identity used in combinatorial proofs	HL Number & Algebra, competition maths
Binomial expansion of (a + b)ⁿ	Binomial distribution P(X = r) = ⁿCᵣ pʳ(1−p)ⁿ⁻ʳ	SL/HL Probability & Statistics
Counting subsets of size r	Multinomial coefficients: choosing r₁, r₂, …, rₖ items simultaneously	University-level combinatorics

How SL 1.9 concepts extend to higher-level topics

Perhaps the most immediately relevant extension for IB students is the binomial distribution in probability. When you run n independent trials, each with probability p of success, the probability of getting exactly r successes is ⁿCᵣ × pʳ × (1 − p)ⁿ⁻ʳ. Notice how the same ⁿCᵣ appears — it counts the number of different ways the r successes can be distributed among the n trials. Mastering combinations now means you'll already have one of the hardest pieces of that topic handled.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain in your own words why ⁿCᵣ is always less than or equal to ⁿPᵣ for the same values of n and r. Under what condition are they equal?

Show answer

Combinations remove duplicate arrangements that permutations count separately. Since ⁿCᵣ = ⁿPᵣ / r!, the combination count is always the permutation count divided by the number of ways to rearrange the chosen items. Because r! ≥ 1, we always have ⁿCᵣ ≤ ⁿPᵣ. The two values are equal only when r! = 1, which happens when r = 0 or r = 1. When you choose 0 or 1 item, there is only one way (or no way) to rearrange the selection, so order is irrelevant.

Question 2: BASIC CALCULATION

Calculate ⁸C₃.

Show answer

Using the formula: ⁸C₃ = 8! / (3! × 5!) = (8 × 7 × 6) / (3 × 2 × 1) = 336 / 6 = 56. You can also verify this using Pascal's Triangle: in row 8, the entry at position r = 3 is 56. A useful shortcut is to write only the 'falling' factors in the numerator (8 × 7 × 6, which is three factors since r = 3) and divide by r! = 3! = 6.

Question 3: INTERMEDIATE

Find the term containing x⁴ in the expansion of (3x − 2)⁶.

Show answer

Set a = 3x, b = −2, n = 6. The general term is ⁶Cᵣ × (3x)⁶⁻ʳ × (−2)ʳ. The power of x is 6 − r, so for x⁴ we need 6 − r = 4, giving r = 2. Then ⁶C₂ = 15. The term is 15 × (3x)⁴ × (−2)² = 15 × 81x⁴ × 4 = 15 × 324x⁴ = 4860x⁴. Note that (−2)² is positive, so the term is +4860x⁴. Always be careful with negative signs — odd powers of a negative b will produce negative terms.

Question 4: APPLIED

A pizza restaurant offers 12 toppings. A customer wants to order a pizza with exactly 4 toppings. How many different pizzas are possible? If the customer's friend orders a pizza with exactly 8 toppings, how many options does the friend have, and why does this make sense?

Show answer

The first customer has ¹²C₄ = 12! / (4! × 8!) = (12 × 11 × 10 × 9) / (4 × 3 × 2 × 1) = 11880 / 24 = 495 possible pizzas. The friend has ¹²C₈ = 12! / (8! × 4!) = 495 possible pizzas — the same number. This illustrates the symmetry property: ¹²C₄ = ¹²C₈ because choosing 4 toppings to include is equivalent to choosing 8 toppings to exclude. Every selection of 4 toppings uniquely determines which 8 are left off, and vice versa.

Question 5: CRITICAL THINKING

The constant term in the expansion of (2x + k/x)⁶ is 4860. Find the possible values of k.

Show answer

Set a = 2x, b = k/x = kx⁻¹, n = 6. The general term is ⁶Cᵣ × (2x)⁶⁻ʳ × (kx⁻¹)ʳ = ⁶Cᵣ × 2⁶⁻ʳ × kʳ × x⁶⁻ʳ × x⁻ʳ = ⁶Cᵣ × 2⁶⁻ʳ × kʳ × x⁶⁻²ʳ. For the constant term, set the exponent of x to zero: 6 − 2r = 0, so r = 3. The constant term is ⁶C₃ × 2³ × k³ = 20 × 8 × k³ = 160k³. Setting this equal to 4860: 160k³ = 4860, so k³ = 4860/160 = 243/8. Therefore k = ∛(243/8) =

(243)^{1/3}

(8)^{1/3}

(3⁵)^{1/3}

/ 2 =

3^{5/3}

/ 2. Since

3^{5/3}

= 3 ×

3^{2/3}

, we get k = 3 × ∛9 / 2. If k can be negative (since k³ = 243/8 has only one real cube root, which is positive), then k = 3∛9 / 2 ≈ 3.12 is the only real solution.

SUMMARY

Lesson Summary

Summary

In this lesson you learned that combinatorics is the mathematics of counting, with roots stretching from ancient India and Persia to Pascal and Newton. The central formula is ⁿCᵣ = n! / (r! × (n − r)!), which counts the number of ways to choose r items from n when order does not matter. You saw that Pascal's Triangle organises all binomial coefficients visually, with each entry being the sum of the two entries above it, and that the triangle exhibits left-right symmetry due to the property ⁿCᵣ = ⁿC₍ₙ₋ᵣ₎.

The binomial theorem expands (a + b)ⁿ as a sum of n + 1 terms, each with the form ⁿCᵣ × aⁿ⁻ʳ × bʳ, where r ranges from 0 to n. To find a specific term or coefficient, determine the correct r value, compute ⁿCᵣ, and remember to raise the entire expressions for a and b (including any internal coefficients) to the appropriate powers. Looking ahead, these same binomial coefficients appear in the binomial probability distribution and connect to generalised expansions in higher-level mathematics.