IB Maths AA HLTopic 1 — Number & AlgebraPaper 1 & 2~9 min read
Binomial Coefficients & Pascal’s Triangle
Right, let’s slow down for a minute. In the previous note, you saw the symbol nCr appear in every binomial expansion. We didn’t pause to ask where it comes from or what it really means — we just used it. This note fixes that. We’ll look at the formula, the four little properties that save you time in exams, and then meet a beautiful pattern called Pascal’s triangle that lets you read off binomial coefficients without any factorials at all. Once you see how Pascal’s triangle and nCr are the same idea wearing different clothes, you’ll start spotting little shortcuts everywhere.
📘 What you need to know
The binomial coefficientnCr = n! ÷ [r!(n − r)!] tells you the number that sits in front of each term in a binomial expansion.
It also has a meaning in counting: nCr = the number of ways to choose r things out of n.
Four useful properties to save time: nC0 = 1, nCn = 1, nC1 = n, and the symmetry rule nCr = nCn−r.
Pascal’s triangle is a stack of numbers where each row starts and ends with 1, and every middle number is the sum of the two directly above it.
The numbers in row n of Pascal’s triangle are exactly nC0, nC1, nC2, …, nCn — so you can read off binomial coefficients without any calculator.
Both the formula for nCr and the binomial theorem are in the formula booklet. Your GDC also computes nCr directly.
For small values of n (say up to 8 or 9), sketching Pascal’s triangle is usually faster than punching factorials into the calculator.
What is a binomial coefficient?
You’ve seen nCr — sometimes written as a vertical pair of numbers in brackets — pop up in every binomial expansion. It’s just a number. A specific kind of number, calculated in a specific way. Here’s the formula:
Binomial coefficientnCr = n!r!(n − r)!✓ in the formula booklet
And remember what the factorial sign means: n! is just n multiplied by every smaller positive integer down to 1. So 5! = 5 × 4 × 3 × 2 × 1 = 120, and 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040. There’s also one weird convention you need: 0! = 1. Don’t fight it; it just works out that way.
Let’s calculate one together. Say we want 6C2. Drop n = 6 and r = 2 into the formula:
Notice the trick: the bottom factorial — here 4! — cancels almost the whole top. So you really only multiply 6 × 5 = 30, then divide by 2! = 2, giving 15. That cancellation always happens. Use it. Don’t multiply out 6! to get 720 if you’re going to immediately divide most of it back out.
Four properties that save you time
Before we move on to Pascal’s triangle, there are four little facts about nCr that turn up over and over in exams. Knowing these means you skip pointless arithmetic.
nC0 = 1
There’s always exactly one way to choose nothing — by doing nothing. So the first number in every row is 1.
nCn = 1
There’s exactly one way to choose all of them — take everything. So the last number in every row is 1.
nC1 = n
Choosing one thing out of n? Easy — there are n options. So the second number in row n is just n.
nCr = nCn−r
Symmetry. Choosing r things to take is the same as choosing n − r things to leave behind.
Why symmetry matters:20C17 looks scary, but it equals 20C3, which is just (20 · 19 · 18) ÷ (3 · 2 · 1) = 1140. Always check whether the easier “mirror” calculation is available before you start.
Meet Pascal’s triangle
Now for the fun part. There’s an absurdly elegant way to find binomial coefficients without using the formula at all. You build a triangle of numbers — called Pascal’s triangle — by following two rules:
🧭 How to build Pascal’s triangle
Every row starts with a 1 and ends with a 1. No exceptions.
Every number in between is the sum of the two numbers directly above it. That’s the entire rule.
That’s it. Two rules, infinite triangle. Let me show you what the first few rows look like, with the addition rule made visible:
Pascal’s triangle — each row is the sum of the two above
See it? The 6 in the middle of row 4 is just 3 + 3 from the row above. The 20 in row 6 is 10 + 10 from row 5. Once you’ve drawn one row, the next one practically writes itself.
A small naming detail. The very top row, with just a single “1”, is called row 0 — it matches n = 0. The next row down (1, 1) is row 1, and so on. So if a question asks for the row corresponding to (a + b)7, you want the row that starts with 1, 7, … — that’s row 7, the eighth one down.
The bridge — Pascal’s triangle = binomial coefficients
Here’s the punchline that ties this whole note together. The numbers in row n of Pascal’s triangle are exactly the binomial coefficients nC0, nC1, nC2, … , nCn. Let me show you with row 4:
So if you want to expand (a + b)4, you don’t need a calculator. Read the row off Pascal’s triangle and stick the powers in:
(a + b)4 = 1·a4 + 4·a3b + 6·a2b2 + 4·ab3 + 1·b4
🤔 Why are Pascal’s triangle and nCr the same?
Because the addition rule of Pascal’s triangle — every entry is the sum of two above — is exactly the same rule that nCr obeys: nCr = n−1Cr−1 + n−1Cr. Both start with the same 1 at the top and follow the same growth rule, so they must produce the same numbers. Pascal’s triangle is just a visual diagram of the formula.
What nCr really counts
One last interpretation that often shows up. The number nCr isn’t only “the coefficient in a binomial expansion” — it’s also the number of ways to choose r objects out of n. So if I have 8 different books on a shelf and I want to grab 3 of them, the number of different combinations I can pick is:
8C3 = 8!3! · 5! = 8 · 7 · 63 · 2 · 1 = 56 ways
🌍 The two faces of nCr
Whenever you see nCr, think of it as wearing two hats: it’s the coefficient in a binomial expansion and it’s a counting answer (“how many ways to choose r from n“). They’re literally the same number — that’s why it’s called a “binomial coefficient” and it’s the same idea you’ll meet again in Permutations & Combinations.
Worked examples
WE 1
Calculate 9C4 by hand
Without using a calculator, find the value of 9C4.
Step 1: Drop n = 9 and r = 4 into the formula9C4 = 9! / [4! · (9 − 4)!] = 9! / (4! · 5!)Step 2: Spot the cancellation — the 5! cancels most of 9!9! = 9 · 8 · 7 · 6 · 5! (write it like this and the 5! divides out)= (9 · 8 · 7 · 6 · 5!) / (4! · 5!)= (9 · 8 · 7 · 6) / 4!Step 3: Multiply top, divide by 4! = 24top: 9 · 8 = 72, 72 · 7 = 504, 504 · 6 = 30243024 ÷ 24 = 1269C4 = 126never multiply 9! out fully — it’s 362880 and pointless. Cancel the bigger factorial first.
WE 2
Use Pascal’s triangle to expand (a + b)7
Find the row of Pascal’s triangle for n = 7 and use it to write the full expansion of (a + b)7.
Step 1: Build down from row 5 — already known: 1, 5, 10, 10, 5, 1row 6: 1, 6, 15, 20, 15, 6, 1 (each entry = sum of two above)row 7: 1, 7, 21, 35, 35, 21, 7, 1Step 2: Stick the powers of a and b on, with a’s dropping and b’s rising(a + b)7 = 1·a7 + 7·a6b + 21·a5b2 + 35·a4b3 + 35·a3b4 + 21·a2b5 + 7·ab6 + b7(a + b)7 = a7 + 7a6b + 21a5b2 + 35a4b3 + 35a3b4 + 21a2b5 + 7ab6 + b7notice the row is symmetric — once you’ve found the first half (1, 7, 21, 35) the rest is mirrored
WE 3
Coefficient of x4 in (1 + x)10
Use Pascal’s triangle to find the coefficient of x4 in the expansion of (1 + x)10.
Step 1: We need 10C4, which is the (4+1)th = 5th entry in row 10build row 8: 1, 8, 28, 56, 70, 56, 28, 8, 1 (from row 7 and the addition rule)row 9: 1, 9, 36, 84, 126, 126, 84, 36, 9, 1row 10: 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1Step 2: 5th entry of row 10 is 210 → that’s 10C4in (1+x)10, the term in x4 is 10C4 · 16 · x4 = 210x4coefficient of x4 = 210first entry is r = 0, so 10C4 is the 5th entry, not the 4th. Off-by-one is the easy slip here.
WE 4
Use symmetry to evaluate 15C12
Without using a calculator, find 15C12.
Step 1: Spot the symmetry — nCr = nCn−r15C12 = 15C15 − 12 = 15C3much easier: choosing 3 to take is the same as choosing 12 to leaveStep 2: Calculate 15C3 directly15C3 = (15 · 14 · 13) / (3 · 2 · 1)top: 15 · 14 = 210, 210 · 13 = 27302730 ÷ 6 = 45515C12 = 455computing 15C12 directly would mean 15! / (12! · 3!) — same answer, but a lot more digits to handle
WE 5
Choosing a committee
A school club has 9 members. In how many different ways can a committee of 4 be chosen from the club?
Step 1: This is “choose r out of n” — straight to nCrn = 9 (total people), r = 4 (committee size)number of ways = 9C4Step 2: Compute (already done in WE 1)9C4 = (9 · 8 · 7 · 6) / 4! = 3024 / 24 = 126there are 126 different committeesorder doesn’t matter — picking Alice then Bob is the same committee as picking Bob then Alice. That’s why it’s nCr not nPr.
WE 6
Pascal’s triangle to expand (2 + x)5
Use Pascal’s triangle to find the full expansion of (2 + x)5.
Step 1: Read off row 5 from Pascal’s trianglerow 5: 1, 5, 10, 10, 5, 1 ← these are the coefficientsStep 2: Powers of 2 drop, powers of x rise — and they sum to 5(2+x)5 = 1·25 + 5·24·x + 10·23·x2 + 10·22·x3 + 5·2·x4 + x5Step 3: Compute each constant= 1·32 + 5·16·x + 10·8·x2 + 10·4·x3 + 5·2·x4 + x5= 32 + 80x + 80x2 + 40x3 + 10x4 + x5(2 + x)5 = 32 + 80x + 80x2 + 40x3 + 10x4 + x5the row 1, 5, 10, 10, 5, 1 is symmetric — but the final answer isn’t, because the powers of 2 break the symmetry
💡 Top tips
Cancel the bigger factorial first. When computing nCr, write n! as n(n−1)…(n−r+1) · (n−r)! and let the (n−r)! cancel. Saves loads of arithmetic.
Check for symmetry before computing.20C18 = 20C2 — much easier. Always look at the smaller of r and n − r.
Pascal’s triangle is fastest for small n. For n ≤ 8 or so, sketching the triangle is quicker than the formula or even the GDC.
Build the triangle row-by-row. Don’t try to jump straight to row 10. Each new row only needs the row above.
Watch the indexing. The first entry of any row is nC0, the second is nC1, and so on. So nCr is the (r+1)th entry, not the rth.
Use your GDC for big values on Paper 2. Look for the “nCr” or “C” button.
Remember the two interpretations.nCr is both a binomial coefficient and a counting answer. Many word problems are nCr in disguise.
⚠ Common mistakes
Off-by-one in Pascal’s triangle. The very top row is row 0, not row 1. The row starting “1, 7, …” is row 7, not row 8.
Multiplying out factorials fully. Computing 9! = 362880 just to divide most of it back out is wasted work. Cancel as you go.
Picking the wrong entry from a row.10C3 is the 4th entry of row 10 (counting from 1), not the 3rd. The first entry corresponds to r = 0.
Forgetting that 0! = 1. Skip this and you’ll get a “divide by zero” panic when computing nC0 or nCn.
Reading the symmetric row but expecting symmetric output. Row 5 is 1, 5, 10, 10, 5, 1 — but (2+x)5 isn’t symmetric in its final form, because the constant 2 raises the early terms.
Adding rows wrong. Each interior entry is the sum of the two directly diagonal above it — not the entries to the side. Watch your alignment when sketching.
Using nCr for problems where order matters. If the question is about arranging or ordering things in a sequence, you want nPr instead. nCr is for unordered selection only.
If this is your first time really seeing Pascal’s triangle, take a few minutes to draw rows 0 through 8 by hand. There’s something about doing it yourself — watching the 1s appear at the edges and the additions falling into place — that locks in the pattern far better than reading about it. From there, every binomial expansion you ever do becomes a quick lookup instead of a calculation. Onwards to the Extension of the Binomial Theorem next, where we’ll see what happens when n is allowed to be a fraction or a negative — and that’s HL-only territory.
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