Choosing a-or-b from each bracket: the combination numbers are the coefficients.
You already know how to square a sum: (a + b)2 = a2 + 2ab + b2. But what about (a + b)5, or (a + b)10? You could multiply it all out by brute force — and lose an afternoon. There is a better way, and it is pure counting in disguise. Expanding (a + b)n means handing out an a or a b from each of the n brackets, and the number in front of each term is just how many ways you could have made that term — a combination number C(n, k). The same numbers live in Pascal’s triangle, where every entry is the sum of the two above it. By the end you will read a coefficient straight off a row, write any single term without expanding, and fish a constant term out of a sixth power in one line.
Start with the smallest case you can still see all of. To expand (a + b)3 means to multiply three copies:
(a + b)(a + b)(a + b)
When you multiply three brackets, every term of the answer is built by reaching into each bracket and grabbing exactly one of its two pieces — an a or a b — and multiplying your three grabs together. One full sweep through all the choices gives all 2 × 2 × 2 = 8 products. List them by how many b’s you took:
Read the picture. There is 1 way to take zero b’s (grab a from all three) → a3. There are 3 ways to take exactly one b (it could come from bracket 1, 2, or 3) → 3a2b. There are 3 ways to take two b’s, and 1 way to take all three. So
(a + b)3 = 1a3 + 3a2b + 3ab2 + 1b3.
The crucial move: a term with exactly k b’s also has n − k a’s (every bracket gives one piece), so it is an−kbk. The only mystery left is the number in front — and that number is just a count of sweeps.
Each term of (a + b)n picks one piece — an a or a b — from every bracket. Take k of the b’s and you must take n − k of the a’s, so the term is an−kbk. The exponents always add to n.
How many ways can you collect a term with exactly k b’s? You are choosing which k of the n brackets hand over a b (the rest hand over an a). That is a selection where order does not matter — choosing brackets {1, 3} is the same term as choosing {3, 1} — so it is a combination, exactly the C(n, k) you built in 32.3 Combinations. The coefficient of an−kbk is
C(n, k) = number of brackets giving a b.
Add up the term for every value of k from 0 to n and you have the whole expansion — the Binomial Theorem:
(a + b)n = Σk=0n C(n, k) an−k bk
= C(n,0)an + C(n,1)an−1b + C(n,2)an−2b2 + … + C(n,n)bn. The numbers C(n,0), C(n,1), …, C(n,n) are called the binomial coefficients.
Check it against the n = 3 case: C(3,0), C(3,1), C(3,2), C(3,3) = 1, 3, 3, 1 — exactly the coefficients we counted. For n = 4 the coefficients are C(4,0), …, C(4,4) = 1, 4, 6, 4, 1, so
(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4.
The coefficient is C(n, k) — the count of brackets that gave a b — not k and not n. In (a + b)4, the a2b2 term has coefficient C(4,2) = 6, never 2 and never 4.
You rarely want the whole expansion — usually you want one specific term. The Binomial Theorem hands you a formula for it. Counting from k = 0, the term carrying bk is the (k + 1)-th term, written
Tk+1 = C(n, k) an−k bk
Set k = 0 for the first term (C(n,0)an), k = 1 for the second, and so on up to k = n for the last (C(n,n)bn).
Why the “+ 1”? Because the b-count k starts at zero, but term numbers start at one. The third term has k = 2, the seventh has k = 6. Keeping this straight is the whole game in the next sections.
Tk+1 carries bk, not bk+1. If a problem asks for the “4th term,” that is k = 3 (since 4 = k + 1), giving C(n,3)an−3b3 — not k = 4.
Dial up n to see the whole expansion appear, then dial k to spotlight one term Tk+1 and watch its coefficient come straight off the lit Pascal row.
Two things to notice as you dial. First, the coefficients of row n really are C(n,0), …, C(n,n) — the lit term’s number sits in the highlighted row at position k. Second, the “add the two above” arrows show where each coefficient came from: C(n, k) is the sum of the two coefficients on its shoulders in row n − 1. That is the Pascal rule, and the next section makes it the star.
Where do the rows come from without computing every C(n, k)? From a rule so simple a child can extend the triangle forever: start and end every row with 1, and fill each inside entry by adding the two numbers above its shoulders.
This “add the two above” rule is not a coincidence — it is the identity you proved in 32.3, made visible:
C(n + 1, k) = C(n, k) + C(n, k − 1).
In words: to build a b-team of size k from n + 1 brackets, either the new bracket gives an a (then choose all k b’s from the old n: C(n, k) ways) or it gives a b (then choose k − 1 more from the old n: C(n, k − 1) ways). “In or out” — add the two cases. That is the two shoulders. So Pascal’s triangle is a table of binomial coefficients, and you never have to multiply out (a + b)n again to find its coefficients — just slide down one more row.
Row 6 of the triangle is 1, 6, 15, 20, 15, 6, 1. So (a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6. No multiplying out — just copy the row.
The binomial coefficients in any row have three habits worth memorizing, and all three follow from the counting story.
Choosing which k brackets give a b is the same as choosing the other n − k to give an a — same split, counted from the other side. So
C(n, k) = C(n, n − k),
and every row reads the same forwards and backwards: 1, 6, 15, 20, 15, 6, 1.
There is only one way to take no b’s and one way to take all of them (the 1’s at the ends), but many balanced ways in between, so the coefficients rise to a peak in the middle and fall back. The bar chart below makes the hill obvious.
Set a = b = 1 in the theorem. The left side is (1 + 1)n = 2n; the right side is just the sum of all the coefficients (every an−kbk becomes 1). So
C(n,0) + C(n,1) + … + C(n,n) = 2n.
That matches the subset count from 32.3: n brackets, each independently giving a or b, makes 2n sweeps in all.
Real problems usually carry numbers inside the brackets — and then the binomial coefficient C(n, k) is only part of the answer. The method is always the same three steps:
1. Write the general term Tk+1 = C(n, k) an−k bk with the actual a and b. 2. Collect the power of x and set its exponent equal to your target. 3. Solve for k, then plug back in.
Here n = 6, a = x, b = 2/x. The general term is
Tk+1 = C(6, k)·x6−k·(2/x)k = C(6, k)·2k·x6−2k.
The constant term has x to the power 0: set 6 − 2k = 0, so k = 3. Then the coefficient is C(6,3)·23 = 20 · 8 = 160. The constant term is 160.
Dial the example below: slide k and watch the x-power of Tk+1 change as 6 − 2k. Land on the power your problem wants and read the full coefficient C(6, k)·2k.
Once a and b carry numbers, the binomial coefficient C(n, k) is not the full coefficient. In (2x − 1)5, the term with x2 takes k = 3 (so the x-power 5 − k = 2), and its full coefficient is C(5,3)·22·(−1)3 = 10 · 4 · (−1) = −40 — not just C(5,3) = 10. The numbers from 2x and from −1 ride along.
Here is a glimpse of where these coefficients lead. Suppose one trial succeeds with chance p and fails with chance q (and p + q = 1). Expand (p + q)n and the term C(n, k) pn−k qk is exactly the chance of getting k failures in n trials — the same C(n, k) counting the ways those k failures could land. That expansion is the binomial distribution, and it opens Stage 33 · Probability. The counting you did here is already the engine of chance.
| Idea | What it says | Why / how |
|---|---|---|
| Binomial Theorem | (a+b)n = Σ C(n,k) an−kbk | each term picks a-or-b from every bracket |
| General term | Tk+1 = C(n,k) an−kbk | b-count k starts at 0 → the “+ 1” |
| Symmetry | C(n,k) = C(n,n−k) | choosing the b’s = choosing the a’s |
| Row sum | ΣC(n,k) = 2n | set a = b = 1 |
| Pascal rule | C(n+1,k) = C(n,k) + C(n,k−1) | sum of the two above; row n = coefficients |
| Specific term | set the x-exponent of Tk+1 = target | solve for k, substitute (numbers ride along) |
Use Pascal’s triangle to expand (a + b)5 fully.
In the expansion of (a + b)8, what is the coefficient of a5b3?
Without adding them one by one, find C(7,0) + C(7,1) + … + C(7,7).
Find the 4th term of (a + b)9.
Find the constant term in the expansion of (x + 2/x)6.
Find the coefficient of x2 in (2x − 1)5.
Six questions to lock it in. Tap the answer you think is right.
This lesson develops CCSS HSA-APR.C.5 — know and apply the Binomial Theorem for the expansion of (x + y)n in powers of x and y with coefficients determined by Pascal’s triangle. It rests directly on HSS-CP.B.9 (use permutations and combinations to compute probabilities and solve problems): the binomial coefficient C(n, k) is the same combination count students met in 32.3, here reappearing as the count of brackets that supply a b. We deliberately tie every coefficient to a counting argument so the formula feels inevitable rather than memorized, and we close by pointing toward the binomial distribution, where these coefficients become probabilities in Stage 33.