Flip a coin five times — in how many ways could you land exactly three heads? Breeding pea plants, dealing card hands, forecasting how often a backup system fails: whenever you count the ways a string of yes/no choices can combine, the same numbers appear. They are the binomial coefficients, and this page is where they come from.

Suppose someone asks you to expand (x + y)^5. You already know how, in principle: multiply every term by every term — see multiplying polynomials. But count what that actually means. (x + y)^5 is five brackets multiplied together, and every one of the five hands you a choice: take the x or take the y. Two choices, five times over — that is 2^5 = 32 separate products to write down, sort, and collect before you're finished. Do it for the seventh power and you're wrangling 128.

Or — you can read the whole answer straight off a triangle of numbers that was already old when Blaise Pascal was born. The triangle isn't a magic trick that happens to work: it is the bookkeeping of all those products, done once, forever, for every power at the same time. This page is about learning to read it.

Start small and watch the pattern appear. Expanding (a + b)^2 the long way gives:

Look only at the numbers in front of each term — the coefficients. They are 1, 2, 1. Expand the next power and the pattern continues:

The coefficients are now 1, 3, 3, 1. Two things are going on in every expansion. First, the powers march in step: the powers of a count down while the powers of b count up, and in every single term they add to n — that part follows the . Second, the coefficients form a pattern of their own. The interesting question is: where do 1, 3, 3, 1 come from — and what comes next?

Why the coefficients count choices

Here's the key insight, and it's worth slowing down for. Write (a + b)^3 as three brackets:

Every product in the expansion is made by picking one letter from each bracket. Pick a all three times and you get a^3 — and there's only one way to do that, so the coefficient of a^3 is 1. But to build a^2b you need the b to come from exactly one bracket — and it could be the first, the second, or the third. Three different routes, all landing on the same term:

Three ways, so the coefficient of a^2b is 3. A binomial coefficient is a count of choices: it counts how many ways you can decide which brackets donate the b.

And that's exactly why Pascal's famous rule — each entry is the sum of the two entries above it — has to be true. Suppose you're choosing which brackets give a b out of n brackets, and look at the last bracket. Either it donates a b (and you finish the job among the other n-1 brackets, needing one fewer b), or it doesn't (and you find all your b's among the other n-1). Two separate piles, no overlap — so the count for n brackets is the sum of two counts for n-1 brackets. That sum is precisely "add the two numbers above." The triangle isn't decoration: every entry is a completed piece of multiplication bookkeeping.

See it built — down to row 6

Pascal's triangle writes the choice-counts out directly. Start with a single 1 at the top (that's row 0 — the triangle counts its rows from zero, matching the power n). Every row begins and ends with 1, and each number inside is just the sum of the two numbers above it. Step through the build below and watch each entry grow from the pair above it — all the way down to row 6.

Read off any row to get the coefficients for that power: row 2 is 1, 2, 1 (that's (a+b)^2), row 3 is 1, 3, 3, 1, and row 5 — 1, 5, 10, 10, 5, 1 — is the answer to our opening challenge, all 32 products of (x+y)^5 collected for you:

A quick sanity check you can always run: the entries of row n add up to 2^n — because they're counting up all 2^n of the raw products. Row 5's entries sum to 1+5+10+10+5+1 = 32. If your row doesn't sum to a power of 2, an addition slipped somewhere. The triangle's diagonals are themselves famous .

Worked example: expand (x + 2)^4

Now put the triangle to work on a binomial where the second term is a number. Here a = x, b = 2 and n = 4, so we take row 4: 1, 4, 6, 4, 1. Lay out the skeleton first — coefficients in order, powers of x descending, powers of 2 ascending:

This is where marks are won and lost: track the powers of the 2 carefully. The 2 isn't a bystander — it climbs through 2^0, 2^1, 2^2, 2^3, 2^4 just as x descends. Evaluate each term:

Notice the final coefficients 1, 8, 24, 32, 16 are not the Pascal row — they're the Pascal row with the powers of 2 multiplied in. The triangle gives you the counting; the powers of b do the rest.

Worked example: expand (2x - 1)^3

Two new wrinkles at once: the first term carries a coefficient (a = 2x), and the second is negative (b = -1). The method doesn't change — but both the coefficient and the sign must ride along inside their brackets, all the way through every power. Row 3 is 1, 3, 3, 1:

Now work each term in full, keeping the brackets until the last moment:

Assemble:

Two things to admire in that answer. The leading term is 8x^3, not 2x^3 — because (2x)^3 cubes the 2 as well as the x. And the signs alternate +,-,+,- — because odd powers of (-1) are negative and even powers are positive. Both facts fall out automatically if you keep the whole of a and the whole of b, sign included, inside the brackets.

Worked example: one coefficient, no full expansion

Exam questions often ask for a single term: say, the coefficient of x^2 in (x + 3)^5. Expanding all six terms would work, but it's five terms of wasted effort. Instead, aim straight at the one you need.

Row 5 is 1, 5, 10, 10, 5, 1, and the general shape of each term is (\text{row entry}) \cdot x^{5-k} \cdot 3^k, the two powers always summing to 5. We want the power of x to be 2, so the 3 must be raised to the power 3 — and the matching row entry (the fourth one, partnering x^2 with 3^3) is 10:

The coefficient of x^2 is 270. One term, three small multiplications, done. This "surgical strike" — pick the row entry, pair it with the right powers, evaluate — is the skill the quiz below will press on hardest.

See it explained

Sal Khan builds Pascal's triangle and uses it to expand binomials — a good second pass at everything above, in a different voice.