Start each row with 1, end it with 1, and make every entry in between the sum of the two directly above it. Row 4 is 1, 4, 6, 4, 1 — and those are exactly the coefficients of (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴.
The rows are binomial coefficients
Multiply out (a + b)³ and you get a³ + 3a²b + 3ab² + b³. The coefficients are 1, 3, 3, 1, which is row 3.
This is not a coincidence, and the reason is a counting argument. Expanding (a + b)ⁿ means choosing, from each of the n brackets, either an a or a b. The coefficient of aⁿ⁻ᵌbᵌ is the number of ways to choose which k brackets contribute the b — which is C(n, k).
So the triangle is a table of “how many ways to choose k things from n”, and the binomial theorem is that table read as coefficients. Row 5 says there are 10 ways to choose 2 things from 5, and it says the coefficient of a³b² is 10, and these are the same statement.
Why every row doubles
Row 0 sums to 1, row 1 to 2, row 2 to 4, row 3 to 8. Row n sums to 2ⁿ, always.
The reason is in the construction. Every entry contributes to exactly two entries in the row below — the one down-left and the one down-right — so the total doubles at every step.
There is a counting reading too. The row sum counts all the subsets of an n-element set: choose none, choose one, choose two, and so on. A set of n things has 2ⁿ subsets, because each element is either in or out. The triangle sorts those subsets by size.
Substituting a = b = 1 into the binomial theorem says the same thing in one line: (1 + 1)ⁿ = 2ⁿ is the sum of the row.
What the diagonals count
First diagonal: all 1s. There is exactly one way to choose nothing.
Second: 1, 2, 3, 4, 5 — the counting numbers. There are n ways to choose one thing from n.
Third: 1, 3, 6, 10, 15 — the triangular numbers. There are n(n−1)/2 ways to choose two things from n, which is also how many handshakes happen in a room of n people.
Fourth: 1, 4, 10, 20, 35 — the tetrahedral numbers, counting the ways to choose three.
There is also the hockey-stick identity: add any diagonal run and the total appears just below and to the side of where you stopped. 1 + 3 + 6 + 10 = 20, and 20 sits at the end of the hook.
Fibonacci hides in it
Read the triangle along its shallow diagonals — not straight down, but at a gentler slope — and each one sums to a Fibonacci number.
1. Then 1. Then 1 + 1 = 2. Then 1 + 2 = 3. Then 1 + 3 + 1 = 5. Then 1 + 4 + 3 = 8. The Fibonacci sequence, from a triangle that was built to count subsets.
The reason is a counting one, as usual: the number of ways to tile a strip of length n with squares and dominoes is F(n+1), and sorting those tilings by how many dominoes they use gives exactly the binomial coefficients along one shallow diagonal.
Counting with it
Committees. How many ways to pick 3 people from 7? Row 7, entry 3 (counting the first as entry 0): 35.
Coin flips. Flip a coin 5 times: row 5 is 1, 5, 10, 10, 5, 1, so there is 1 way to get no heads, 5 ways to get one, 10 ways to get two. Divide by the row sum of 32 and you have the probabilities.
Paths on a grid. The number of shortest routes to a point on a lattice is the entry at that position, for the same reason the triangle is built the way it is: you arrive from one of two places.
The symmetry. C(n, k) = C(n, n−k), so every row reads the same backwards. Choosing 3 people to include is the same as choosing 4 to leave out.
Its history
Pascal wrote about it in 1654, and the name stuck in the West. He was several centuries late.
The Persian mathematician Al-Karaji described it around 1000 CE, and Omar Khayyam worked with it shortly after — it is still called the Khayyam triangle in Iran. In China, Jia Xian had it by about 1100 and Yang Hui published it in 1261, which is why it is Yang Hui’s triangle there. Indian scholars knew the underlying combinatorics earlier still, in Pingala’s work on prosody.
What Pascal added was the systematic treatment, and the connection to probability that he developed with Fermat. The triangle was old; using it to reason about chance was not.
Sources and methodology
The triangle is standard, and its history is genuinely older than the name; these are the references.
Method. Each row is built from the one above by the multiplicative recurrence C(n, k) = C(n, k−1)(n − k + 1)/k on arbitrary-precision integers, which never forms a factorial and never divides inexactly — every intermediate is itself a binomial coefficient. The suite checks all 351 entries up to row 25 against the factorial definition computed separately, and asserts that every row to 40 sums to exactly 2ⁿ and that each entry equals the two above it. That engine is verified on every change against 76 hand-written assertions, including that every entry to row 40 equals the sum of the two above it and every row sums to exactly 2ⁿ — a pair of checks an arithmetic slip anywhere in the triangle would fail. The count and the per-case breakdown are published on the formula verification page.
Read the guide
For a single coefficient at a row far past 40, the Binomial Coefficient Calculator computes C(n, k) directly. The shallow diagonals of this triangle sum to the Fibonacci numbers, which the Fibonacci Calculator takes further.