Pascal's Triangle - Definition, Patterns, Examples

TL;DR
Pascal's Triangle is a triangular array where every number is the sum of the two directly above it, and each row holds the coefficients of a binomial expansion. This article shows how to build the triangle, the patterns hiding inside it, the link to (\binom{n}{r}), and the mistakes students make first.

A Shape That Predicts Every Coin Toss Before You Flip

One triangle of numbers already knows the odds of every coin toss you will ever make. The row that starts 1, 5, 10, 10, 5 tells you there are ten ways out of thirty-two to get three heads in five flips. That is Pascal's Triangle doing its quiet work.

What Is Pascal's Triangle?

Pascal's Triangle is a triangular arrangement of numbers in which the number at the top is 1, every number on the slanted edges is 1, and every interior number equals the sum of the two numbers diagonally above it. The rows are numbered from (n=0) at the top, and the entries across row (n) are exactly the coefficients you get when you expand ((a+b)^n).

That second fact is why the triangle earns its place in algebra. Write out ((a+b)^2 = a^2 + 2ab + b^2) and the coefficients are 1, 2, 1, row 2. Expand ((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3) and the coefficients are 1, 3, 3, 1, row 3. The triangle hands you the numbers without any multiplying out.

How Do You Build Pascal's Triangle?

Start with a single 1 at the apex. That is row 0. Each new row begins and ends with 1, and you fill the inside by adding neighbors from the row above.

Building row 4 from row 3 (1, 3, 3, 1):
1+3=4; 3+3=6; 3+1=4 → Row 4 is 1, 4, 6, 4, 1. The rule is sometimes written as Pascal's rule:
(\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r})
Here (\binom{n}{r}) is the entry in row (n), position (r) (counting positions from 0). It is the same as the binomial coefficient, computed directly by
(\binom{n}{r} = \frac{n!}{r!(n-r)!})
where (n! = n × (n-1) × ⋯ × 2 × 1). You never need the formula to build the triangle — addition alone does it — but the formula lets you jump straight to a single entry without writing every row first.

The Patterns Hidden Inside Pascal's Triangle

The triangle is famous because so many number patterns live inside one simple rule. The patterns most worth knowing:

• Row sums are powers of 2. Add every entry in row (n) and you get (2^n). Row 3: 1+3+3+1=8=2^3.
• The triangle is symmetric. Each row reads the same left-to-right as right-to-left, because (\binom{n}{r} = \binom{n}{n-r}).
• The diagonals count things. The first diagonal is all 1s; the second is the counting numbers 1, 2, 3, 4,…; the third is the triangular numbers 1, 3, 6, 10,….
• The shallow diagonals add to Fibonacci numbers. Sum the entries along a gentle diagonal slice and you get 1, 1, 2, 3, 5, 8,…, the Fibonacci sequence.
• Binomial coefficients. Row (n) is the coefficient list for ((a+b)^n), the property that ties the triangle to the binomial theorem.

Not every pattern is on this list — there are more, and you will meet them later. These five carry most of the weight at this level.

Examples of Pascal's Triangle

Example 1

Write out the first five rows of Pascal's Triangle.
Start at row 0 and apply the add-the-two-above rule:

• Row 0: 1
• Row 1: 1 1
• Row 2: 1 2 1
• Row 3: 1 3 3 1
• Row 4: 1 4 6 4 1

Final answer: the rows are 1; 1 1; 1 2 1; 1 3 3 1; 1 4 6 4 1.

Example 2

Use Pascal's Triangle to expand ((a+b)^4).
The coefficients from row 4 are 1, 4, 6, 4, 1. Thus,
((a+b)^4 = 1a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + 1b^4).

Final answer: ((a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4).

Example 3

Find the entry in row 6, position 2.
Using the formula:
(\binom{6}{2} = \frac{6!}{2!4!} = 15).

Final answer: (\binom{6}{2} = 15).

Example 4

Expand ((x+2)^3) using Pascal's Triangle.
Row 3 gives the coefficients 1, 3, 3, 1.
((x+2)^3 = 1x^3 + 3x^2(2) + 3x(2^2) + 1(2^3) = x^3 + 6x^2 + 12x + 8).

Final answer: ((x+2)^3 = x^3 + 6x^2 + 12x + 8).

Example 5

A fair coin is tossed five times. In how many ways can exactly two heads appear?
The number of ways to choose which 2 of the 5 tosses are heads is:
(\binom{5}{2} = \frac{5!}{2!3!} = 10).

Final answer: 10 ways.

Example 6

Confirm that the entries in row 4 sum to 16.
Row 4 is 1, 4, 6, 4, 1.

1 + 4 + 6 + 4 + 1 = 16 = 2^4.

Final answer: the row-4 sum is 16 = 2^4.

Why the Triangle Keeps Reappearing

Pascal's Triangle survives because it answers a question humans keep asking: in how many ways can things combine?

Where it earns its keep:

• Probability. Tossing coins, drawing cards, modelling errors — the rows of the triangle are the binomial distribution.
• Algebra. Expanding ((a+b)^n) by hand is slow; the triangle reads off the coefficients instantly.
• Finance and computing. The binomial option-pricing model and many recursive algorithms lean on the same add-the-two-above structure.

Where Students Trip Up on Pascal's Triangle

Mistake 1: Counting rows from one instead of zero

Correct way: Treating row 2 as the row for ((a+b)^2) corresponds to counting down to "the nth row I wrote."

Mistake 2: Forgetting the descending and ascending powers

Correct way: Each term carries both variables.

Mistake 3: Misreading positions in the choose formula

Correct way: Always write the full denominator.

Build It Before You Memorise the Formula

Introducing Pascal's Triangle by hand before memorising the formula means students see why the coefficients are what they are.

Conclusion

• Pascal's Triangle is a number array where each interior entry is the sum of the two directly above it.
• Row (n) holds the coefficients of ((a+b)^n) and equals the binomial coefficients (\binom{n}{r}).
• Its built-in patterns include power-of-two row sums, left-right symmetry, and Fibonacci numbers on shallow diagonals.
• The most common mistakes are counting rows from 1, dropping a variable, and misreading positions.

Practice These to Solidify Your Understanding

1. Write out rows 0 through 6 of Pascal's Triangle.
2. Expand ((x+1)^5) using the correct row.
3. Compute (\binom{8}{3}) with the choose formula.