Pascal’s triangle

What just happened?

Every number in the triangle is the sum of the two numbers directly above it (the outermost numbers in each row are always 1). This simple addition rule generates, row by row, exactly the same numbers you get from the binomial-coefficient formula — this is no coincidence, they are two sides of the same coin.

A little history

This triangle was known to mathematicians in China (Yang Hui, around 1261), in Persia and in India long before Europe. But it was the French mathematician and philosopher Blaise Pascal who, in his treatise “Traité du triangle arithmétique” — written in 1654 and published posthumously in 1665 — systematised its properties and connected them with probability and combinatorics. Earlier, corresponding with Pierre de Fermat on problems of games of chance, Pascal laid the foundations of probability theory.

Binomial coefficient:

— the number of ways to choose k elements from an n-element set (without repetition, order irrelevant). Row n of the triangle is, in order: $C(n,0), C(n,1), \dots, C(n,n)$.

The binomial theorem:

That is why row n of the triangle is exactly the set of coefficients in the expansion of $(a+b)^n$ — e.g. for n = 4: $a^4+4a^3b+6a^2b^2+4ab^3+b^4$, and the numbers 1, 4, 6, 4, 1 are precisely row 4.

Hidden patterns in the diagonals

The outermost diagonal is all ones, the next one the natural numbers (1, 2, 3, 4…), and the third (highlighted in yellow) the triangular numbers: 1, 3, 6, 10, 15… The sum of all the numbers in row n is always exactly $2^n$ — this is the invariant of this model.

Why this matters for your exams:

Problems about choosing k elements from n (combinations), the binomial theorem and sums of coefficients are all solved by the same triangle. The trick: don’t compute factorials by hand for small n — read the value straight off the triangle.

Remember: every term is the sum of the two above it, row n holds the coefficients of $(a+b)^n$, and the sum of row n is always $2^n$.