How to build the triangle
Write 1 at the top. This is row 0. Row 1 is 1 1. For every row after that, put a 1 at each end. Each number in the middle is the sum of the two numbers just above it.
Row 2: 1, 1+1, 1 = 1 2 1. Row 3: 1, 1+2, 2+1, 1 = 1 3 3 1. Row 4: 1 4 6 4 1. Row 5: 1 5 10 10 5 1. Row 6: 1 6 15 20 15 6 1. Row 7: 1 7 21 35 35 21 7 1.
The edge numbers are 1 because an edge number has only one number above it, and that number is 1.
Patterns hiding in the triangle
- Mirror: each row reads the same from left to right and from right to left.
- Row sums: 1, 2, 4, 8, 16, 32… Row n adds up to 2ⁿ. Each number is passed to two numbers below, so the total doubles.
- Slanted lines: the first slanted line is all 1s, the next is 1, 2, 3, 4… (counting numbers) and the next is 1, 3, 6, 10, 15… (triangular numbers).
- Powers of 11: the first rows give 11⁰ = 1, 11¹ = 11, 11² = 121, 11³ = 1331 and 11⁴ = 14641.
Row n gives the coefficients of (a + b)ⁿ
Multiply out: (a + b)² = a² + 2ab + b². The numbers in front are 1 2 1, which is row 2. Multiply again by (a + b): (a + b)³ = a³ + 3a²b + 3ab² + b³. The numbers are 1 3 3 1, row 3.
Why? When you multiply by (a + b) once more, each new term comes from two old terms: one times a and one times b. So each new coefficient is the sum of two old ones. This is the same adding rule as the triangle.
To expand (a + b)ⁿ: take row n. The power of a goes down from n to 0, the power of b goes up from 0 to n, and the powers in each term add up to n. There are n + 1 terms.
For (a + b)⁴ the row is 1 4 6 4 1: a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴.
For (a − b)ⁿ use the same numbers but the signs go + − + − …: (a − b)³ = a³ − 3a²b + 3ab² − b³. For numbers other than 1, put them in place of a or b, e.g. (x + 2)³ = x³ + 3·x²·2 + 3·x·2² + 2³ = x³ + 6x² + 12x + 8.
One triangle, many names
Yang Hui, a Chinese mathematician of the 13th century, showed this triangle in his book and said it came from an earlier scholar, Jia Xian. That is why it is called Yang Hui's triangle in China. In India it is known as Meru Prastara (the steps of the Mount Meru) from the study of poetry rhythms by Pingala and later by Halayudha. Persian poets and mathematicians knew it too. In Europe it is called Pascal's triangle after Blaise Pascal, who wrote about it in the 17th century. The idea was found in many places over a thousand years.
Try it: coins and paths
Coin trick: Toss 4 coins together 32 times and count how many heads come up each time. Most throws will show 2 heads, and 0 or 4 heads will be rare. The triangle row 1 4 6 4 1 tells you the chances: 6 of the 16 outcomes have 2 heads, and only 1 has 4 heads.
Path counting: Put a pebble on the top block. A pebble can go down-left or down-right. The number on each block is the number of different ways to reach it from the top. Check this for the block 6 in row 4 on a drawing.
In the 3D above: slide n and predict the row before you look.
Key formulas and definitions
- Inside number = the two numbers above it added together; edge numbers = 1
- Row n has n + 1 numbers and adds up to 2ⁿ
- (a + b)ⁿ: coefficients from row n; power of a goes n, n−1, …, 0 and power of b goes 0, 1, …, n
- (a − b)ⁿ: same coefficients, signs alternate + − + −
- Rows: 1 | 1 1 | 1 2 1 | 1 3 3 1 | 1 4 6 4 1 | 1 5 10 10 5 1 | 1 6 15 20 15 6 1
Worked examples
1. Write row 5 using row 4 (1 4 6 4 1).
Put 1 at each end. In between: 1 + 4 = 5, 4 + 6 = 10, 6 + 4 = 10, 4 + 1 = 5. Row 5 is 1 5 10 10 5 1.
2. Expand (a + b)⁴.
Row 4 is 1 4 6 4 1. So (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴. Check: the powers in each term add up to 4.
3. Expand (x + 1)⁵.
Row 5 is 1 5 10 10 5 1. The powers of 1 are all 1, so (x + 1)⁵ = x⁵ + 5x⁴ + 10x³ + 10x² + 5x + 1.
4. Expand (a − b)³.
Row 3 is 1 3 3 1. Signs go + − + −. So (a − b)³ = a³ − 3a²b + 3ab² − b³.
5. Expand (x + 2)³.
Row 3 is 1 3 3 1 with a = x and b = 2. Terms: 1·x³ = x³; 3·x²·2 = 6x²; 3·x·4 = 12x; 1·8 = 8. So (x + 2)³ = x³ + 6x² + 12x + 8.
6. Find the sum of the numbers in row 7, and check by adding. How many terms has (a + b)⁹?
Row 7 adds up to 2⁷ = 128. Check: 1 + 7 + 21 + 35 + 35 + 21 + 7 + 1 = 128. (a + b)⁹ has 9 + 1 = 10 terms.
7. Use the triangle to find 11⁴, and to find the coefficient of x³ in (x + 1)⁶.
Row 4 is 1 4 6 4 1, and these are single digits, so 11⁴ = 14641. Row 6 is 1 6 15 20 15 6 1. The terms go x⁶, x⁵, x⁴, x³ … so x³ is the fourth term and its coefficient is 20.
Common mistakes
- Counting the top 1 as row 1. The top row is row 0, so row n is the one that gives (a + b)ⁿ.
- Forgetting the 1 at each end when building a new row, or adding three numbers instead of two.
- Writing the powers wrongly. In every term the powers of a and b must add up to n.
- Forgetting the alternate signs in (a − b)ⁿ, or putting numbers into a or b without raising them to the right power.