Pascal's Triangle: A Shortcut to Combinations, Binomial Expansions, and Probabilities

Building the Triangle

Start with 1. Each subsequent row begins and ends with 1, and each interior number is the sum of the two above. Row 5: 1, 5, 10, 10, 5, 1. Row 6: 1, 6, 15, 20, 15, 6, 1. Row 10: 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1. For example, 45 + 120 = 165 appears in row 11. This addition rule corresponds to the identity C(n, k) = C(n−1, k−1) + C(n−1, k): to choose k items from n, either include a particular item (and choose k−1 from the rest) or exclude it (and choose k from the rest).

The numbers are symmetric: C(n, k) = C(n, n−k), because choosing which k to include is the same as choosing which n−k to leave out.

Binomial Expansion

The entries of row n are the coefficients in the expansion of (a + b)^n. For n = 4: (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴, with coefficients 1, 4, 6, 4, 1. For n = 5, (x + 2)⁵ = x⁵ + 5·2x⁴ + 10·4x³ + 10·8x² + 5·16x + 32 = x⁵ + 10x⁴ + 40x³ + 80x² + 80x + 32. This avoids multiplying out repeated binomials and helps with algebra, probability, and series approximations, such as (1 + x)ⁿ ≈ 1 + nx for small x.

Counting and Probability

The row sums equal 2ⁿ: the total number of subsets of an n-element set. For 10 items, there are 1,024 subsets, of which 252 contain exactly 5 items. For coin flips, the probability of exactly k heads in n flips of a fair coin is C(n, k) ÷ 2ⁿ. With 10 flips, P(5 heads) = 252 ÷ 1,024 = 24.6%, P(exactly 3) = 120 ÷ 1,024 = 11.7%, and P(8 or more) = (45 + 10 + 1) ÷ 1,024 = 5.5%. A committee of 3 chosen from 10 people can be formed in 120 ways, which you can read straight from row 10, position 3.

The number of paths on a grid from one corner to the opposite corner using only steps right and down is a Pascal number. A 4 × 3 grid, which needs 4 rights and 3 downs, has C(7, 3) = 35 paths.

Patterns Hidden in the Triangle

The diagonal next to the edge contains the counting numbers 1, 2, 3, 4...; the next diagonal has triangular numbers 1, 3, 6, 10, 15; the next has tetrahedral numbers 1, 4, 10, 20, 35. Summing the diagonals at a slant produces the Fibonacci numbers 1, 1, 2, 3, 5, 8. Coloring the odd numbers produces the Sierpinski triangle fractal. Row n of the triangle, for n = 0 to 4, gives the digits of powers of 11: 11⁴ = 14,641.

These patterns make Pascal's triangle a favorite for exploring number theory.

When to Use It

For small values of n up to about 15, writing out the triangle is faster than using factorials. For larger values, use the combination formula or a calculator, since the numbers grow quickly: C(52, 5) = 2,598,960, the number of poker hands.