# Pascal's triangle: patterns, coin flips and the numbers behind combinations

> Start with a single 1 and add two numbers to get the one below. That tiny rule builds a triangle hiding powers of 2, the Fibonacci sequence, the coefficients of (a + b)ⁿ and the odds of every coin-flip result. This guide walks through each pattern with pictures.

Published 2026-10-09 by Toolcloak. 4 minute read. Original: https://toolcloak.pages.dev/articles/pascals-triangle-patterns-coin-flips-and-the-numbers-behind-combinations
Related free tool: [Pascal’s Triangle Generator](https://toolcloak.pages.dev/tools/pascals-triangle-generator)

**Pascal's triangle** is built with one rule: every number is the sum of the two numbers directly above it, and the edges are all 1. It is named after the French mathematician Blaise Pascal, but it was known in India, Persia and China centuries earlier.

> **In short**
> - Row n (counting from 0) lists the **combinations C(n, k)**: the number of ways to pick k things from n.
> - Each row adds up to **2ⁿ**: 1, 2, 4, 8, 16, 32…
> - Row n gives the coefficients in the expansion of **(a + b)ⁿ**.
> - Row 10 gives the exact odds of getting 0, 1, 2… 10 heads in ten coin flips.

## The first nine rows

| Row | Numbers |
| --- | --- |
| 0 | 1 |
| 1 | 1  1 |
| 2 | 1  2  1 |
| 3 | 1  3  3  1 |
| 4 | 1  4  6  4  1 |
| 5 | 1  5  10  10  5  1 |
| 6 | 1  6  15  20  15  6  1 |
| 7 | 1  7  21  35  35  21  7  1 |
| 8 | 1  8  28  56  70  56  28  8  1 |

*To get the 6 in row 4, add the 3 and 3 above it. To get the 56 in row 8, add 21 and 35 from row 7.*

**Building row 5 from row 4**

1. **Start with row 4.** 1  4  6  4  1
2. **Put a 1 at each end.** 1  _  _  _  _  1
3. **Add neighbours in pairs.** 1+4 = 5, 4+6 = 10, 6+4 = 10, 4+1 = 5
4. **Row 5 is ready.** 1  5  10  10  5  1

*No multiplication is needed. Repeated addition produces every row.*

Try it: [Pascal’s Triangle Generator](https://toolcloak.pages.dev/tools/pascals-triangle-generator). Generate as many rows as you like and highlight patterns such as odd numbers and sums.

## Pattern 1: each row sums to a power of 2

Add up any row and you get 2 raised to the row number. Row 4: 1 + 4 + 6 + 4 + 1 = 16 = 2⁴. The reason is simple: each number feeds two numbers below it, so each row doubles the one before.

**Sum of each row (rows 0 to 10)**

|  | Value |
| --- | --- |
| 0 | 1 |
| 1 | 2 |
| 2 | 4 |
| 3 | 8 |
| 4 | 16 |
| 5 | 32 |
| 6 | 64 |
| 7 | 128 |
| 8 | 256 |
| 9 | 512 |
| 10 | 1024 |

*A perfect doubling staircase. Row 10 adds up to 1,024.*

## Pattern 2: the numbers are combinations

The number in row **n**, position **k** (both counted from 0) is **C(n, k)**, the count of ways to choose k items from n without caring about order. In row 5, the 10 is C(5, 2): there are 10 ways to pick 2 people from 5.

**C(n, k) for small n and k**

| n \ k | 0 | 1 | 2 | 3 | 4 | 5 |
| --- | --- | --- | --- | --- | --- | --- |
| n = 2 | 1 | 2 | 1 |  |  |  |
| n = 3 | 1 | 3 | 3 | 1 |  |  |
| n = 4 | 1 | 4 | 6 | 4 | 1 |  |
| n = 5 | 1 | 5 | 10 | 10 | 5 | 1 |
| n = 6 | 1 | 6 | 15 | 20 | 15 | 6 |
| n = 7 | 1 | 7 | 21 | 35 | 35 | 21 |

Real examples: choosing a 5-card hand from 52 cards has C(52, 5) = **2,598,960** possibilities, and an 6-from-49 lottery has C(49, 6) = **13,983,816** possible tickets. The triangle gets there by repeated addition, and a formula does it in one step.

## Pattern 3: coin flips

Flip a fair coin four times. There are 2⁴ = 16 equally likely sequences, and the number of those with exactly k heads is C(4, k): row 4 of the triangle.

**Ways to get k heads in 4 flips (out of 16 sequences)**

|  | Value |
| --- | --- |
| 0 heads | 1 |
| 1 head | 4 |
| 2 heads | 6 |
| 3 heads | 4 |
| 4 heads | 1 |

*Two heads and two tails is the most common result, with 6 of 16 sequences (37.5%). All heads has just 1 of 16 (6.25%).*

Move to ten flips and row 10 gives the counts out of 1,024. The bell shape appears: results near 5 heads are far more common than extremes.

**Ways to get k heads in 10 flips (out of 1,024 sequences)**

|  | Value |
| --- | --- |
| 0 | 1 |
| 1 | 10 |
| 2 | 45 |
| 3 | 120 |
| 4 | 210 |
| 5 | 252 |
| 6 | 210 |
| 7 | 120 |
| 8 | 45 |
| 9 | 10 |
| 10 | 1 |

*Exactly 5 heads happens 252 times out of 1,024, about 24.6%. All heads or all tails happens 2 times out of 1,024, about 0.2%.*

**Odds in 10 fair flips**

|  | Value | Note |
| --- | --- | --- |
| Exactly 5 heads | 24.6% | 252 of 1,024 |
| 4, 5 or 6 heads | 65.6% | 672 of 1,024 |
| Outside 4 to 6 | 34.4% | 352 of 1,024 |
| Exactly 10 heads | 0.1% | 1 of 1,024 |

*Add the middle three counts, 210 + 252 + 210, to get 672. Even a fair coin gives lopsided results more often than most people expect.*

## Pattern 4: expanding (a + b)ⁿ

Multiply (a + b) by itself n times and the numbers in front of each term are row n of the triangle. For (a + b)⁴ the coefficients are 1, 4, 6, 4, 1.

| Power | Expansion |
| --- | --- |
| (a + b)² | a² + 2ab + b² |
| (a + b)³ | a³ + 3a²b + 3ab² + b³ |
| (a + b)⁴ | a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴ |
| (a + b)⁵ | a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵ |

*Set a = b = 1 and every coefficient adds to 2ⁿ, which is the row-sum pattern again.*

## Pattern 5: hidden Fibonacci numbers

Draw shallow diagonals through the triangle and add the numbers on each one. The sums are 1, 1, 2, 3, 5, 8, 13: the **Fibonacci sequence**. For example, 1 + 6 + 5 + 1 = 13 on one such diagonal.

**Diagonal sums spell out Fibonacci**

| When | What |
| --- | --- |
| Sum 1 | 1. A single 1. |
| Sum 2 | 1 + 1 = 2. Two 1s from neighbouring rows. |
| Sum 5 | 1 + 3 + 1 = 5. Picked from rows 2, 3 and 4. |
| Sum 8 | 1 + 4 + 3 = 8. Picked from rows 3, 4 and 5. |
| Sum 13 | 1 + 5 + 6 + 1 = 13. Picked from rows 3, 4, 5 and 6. |

*Fibonacci numbers: each is the sum of the two before it. The same numbers sit hidden inside the triangle.*

## More patterns to spot

**Quick pattern guide**

| Box | What it means |
| --- | --- |
| Hockey stick | Add a diagonal run from the edge, such as 1 + 3 + 6 + 10 = 20, and the next number down and across is the sum. |
| Triangular numbers | The third diagonal reads 1, 3, 6, 10, 15, 21: the number of dots in a triangle. |
| Powers of 11 | Rows 0 to 4 read as 1, 11, 121, 1331, 14641, which are 11⁰ to 11⁴. |
| Odd and even | Colour odd numbers and a fractal called the Sierpinski triangle appears. |

*From row 5 the entries reach 10 or more and carry over to the next place: row 5 (1, 5, 10, 10, 5, 1) becomes 161,051, which is 11⁵.*

## Frequently asked questions

### What is Pascal's triangle?

It is a triangle of numbers where each number is the sum of the two directly above it, with 1s along both edges. Its rows contain the binomial coefficients.

### How do you find a number in Pascal's triangle?

Add the two numbers above it. Or use the formula C(n, k) = n! ÷ (k! × (n − k)!) for the number in row n, position k, counting both from zero.

### What do the row sums of Pascal's triangle equal?

Each row sums to 2 raised to the row number. Row 5 adds up to 32 and row 10 adds up to 1,024.

### How is Pascal's triangle used in probability?

Row n gives the number of ways to get k successes in n fair yes-or-no trials, such as k heads in n coin flips. Divide each entry by 2ⁿ to get its probability.

### Where is the Fibonacci sequence in Pascal's triangle?

Add the numbers along the shallow diagonals. The sums are 1, 1, 2, 3, 5, 8, 13 and so on.

### Who invented Pascal's triangle?

Blaise Pascal wrote about it in the 1650s, but earlier mathematicians in India, Persia and China studied it long before. In China it is often called Yang Hui's triangle.