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 $1, 5, 10, 10, 5, 1$ 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 neighbours from the row above.
Building row $4$ from row $3$ ($1, 3, 3, 1$):
$$1$$ $$1 + 3 = 4$$ $$3 + 3 = 6$$ $$3 + 1 = 4$$ $$1$$
So 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 $\binom{n}{r}$ (read "$n$ choose $r$") 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 $n!$ ("$n$ factorial") means $n \times (n-1) \times \cdots \times 2 \times 1$. 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 $1$s; the second is the counting numbers $1, 2, 3, 4, \ldots$; the third is the triangular numbers $1, 3, 6, 10, \ldots$
The shallow diagonals add to Fibonacci numbers. Sum the entries along a gentle diagonal slice and you get $1, 1, 2, 3, 5, 8, \ldots$, 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 (the hockey-stick identity, the prime-row divisibility rule), 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:
$$\text{Row } 0: \quad 1$$ $$\text{Row } 1: \quad 1 \quad 1$$ $$\text{Row } 2: \quad 1 \quad 2 \quad 1$$ $$\text{Row } 3: \quad 1 \quad 3 \quad 3 \quad 1$$ $$\text{Row } 4: \quad 1 \quad 4 \quad 6 \quad 4 \quad 1$$
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 $(a + b)^4$.
A natural first move is to grab the row you want by counting rows from 1. Say you reach for the fourth row written down, $1, 3, 3, 1$, and write $(a+b)^4 = a^4 + 3a^3b + 3a^2b^2 + b^3$. Check the powers: the exponents on $a$ and $b$ in each term should add to $4$, but $a^3b$ gives $3 + 1 = 4$ while the last term $b^3$ gives only $3$. The expansion is short a term, and the coefficients are wrong.
The slip is counting the rows from $1$ instead of from $0$. The coefficients of $(a+b)^4$ come from row $4$, which is the fifth row written down: $1, 4, 6, 4, 1$.
$$(a+b)^4 = 1a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + 1b^4$$
The power of $a$ falls from $4$ to $0$ while the power of $b$ climbs from $0$ to $4$, and every term's exponents sum to $4$.
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 $2$ (counting positions from $0$).
Using the formula:
$$\binom{6}{2} = \frac{6!}{2!,(6-2)!} = \frac{6!}{2!,4!}$$ $$= \frac{6 \times 5}{2 \times 1} = \frac{30}{2} = 15$$
Row $6$ is $1, 6, 15, 20, 15, 6, 1$, and position $2$ is indeed $15$.
Final answer: $\binom{6}{2} = 15$.
Example 4
Expand $(x + 2)^3$ using Pascal's Triangle.
Row $3$ gives the coefficients $1, 3, 3, 1$. Apply them with $a = x$ and $b = 2$:
$$(x+2)^3 = 1\cdot x^3 + 3\cdot x^2(2) + 3\cdot x(2)^2 + 1\cdot(2)^3$$ $$= x^3 + 6x^2 + 12x + 8$$
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 the entry in row $5$, position $2$:
$$\binom{5}{2} = \frac{5!}{2!,3!} = \frac{5 \times 4}{2 \times 1} = 10$$
Row $5$ reads $1, 5, 10, 10, 5, 1$, and position $2$ is $10$. So there are $10$ ways out of $2^5 = 32$ total outcomes.
Final answer: $10$ ways.
Example 6
Confirm that the entries in row $4$ sum to $2^4$.
Row $4$ is $1, 4, 6, 4, 1$.
$$1 + 4 + 6 + 4 + 1 = 16 = 2^4$$
The row sum matches the power-of-two pattern.
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? The Persian poet-mathematician Omar Khayyam and the Chinese mathematician Yang Hui both wrote down the array centuries before Blaise Pascal — it kept being rediscovered precisely because the counting problem it solves is everywhere.
Where it earns its keep:
Probability. Tossing coins, drawing cards, modelling errors — the rows of the triangle are the binomial distribution. The row-$5$ entries told us, with no extra work, how three-heads-in-five splits across all outcomes.
Algebra. Expanding $(a+b)^n$ by hand is slow; the triangle reads off the coefficients instantly, which is why it underpins the binomial theorem.
Finance and computing. The binomial option-pricing model and many recursive algorithms lean on the same add-the-two-above structure.
The destination here is the binomial theorem: once you trust that row $n$ holds the coefficients of $(a+b)^n$, you can expand a power without multiplying anything out — and that shortcut scales to expansions far too large to draw a triangle for.
Where Students Trip Up on Pascal's Triangle
Mistake 1: Counting rows from one instead of zero
Where it slips in: When a problem asks for the coefficients of $(a+b)^n$ and the student counts down to "the $n$th row I wrote."
Don't do this: Treating $1, 2, 1$ as the row for $(a+b)^3$ because it is the third line on the page.
The correct way: The apex is row $0$. Row $2$ ($1, 2, 1$) belongs to $(a+b)^2$; row $3$ ($1, 3, 3, 1$) belongs to $(a+b)^3$. Counting from $0$ aligns the row index with the exponent.
Mistake 2: Forgetting the descending and ascending powers
Where it slips in: When writing the expansion after correctly reading off the coefficients.
Don't do this: Writing $(a+b)^4 = a^4 + 4a^3 + 6a^2 + 4a + 1$, attaching the coefficients to powers of $a$ alone and dropping $b$.
The correct way: Each term carries both variables: the power of $a$ falls from $n$ to $0$ while the power of $b$ rises from $0$ to $n$, and the two exponents always add to $n$. The most common version of this error is dropping the middle term entirely — the $a^4 + b^4$ habit — exactly the kind of slip that defining "every term's exponents sum to $n$" is meant to catch before it happens.
Mistake 3: Misreading positions in the choose formula
Where it slips in: When jumping to a single entry with $\binom{n}{r}$ instead of building the row.
Don't do this: Computing $\binom{6}{2}$ as $\frac{6!}{2!}$ and forgetting the $(n-r)!$ in the denominator, which gives $360$ instead of $15$.
The correct way: Always write the full denominator $r!,(n-r)!$. The skipped $(n-r)!$ is the single quietest source of wrong binomial coefficients, and it never shows up if you write the formula out in full before substituting.
A real version of this counting blind spot shows up outside the classroom. Early lottery and insurance miscalculations came from undercounting the number of ways outcomes can combine — exactly what the triangle's entries measure. The British actuary tables of the eighteenth century leaned on binomial-coefficient reasoning to price risk correctly; getting the combination count wrong meant mispricing real money, which is why binomial-coefficient methods became foundational to actuarial science.
Build It Before You Memorise the Formula
Introducing Pascal's Triangle by hand before the binomial-coefficient formula means students see why the coefficients are what they are rather than memorising $\binom{n}{r}$ cold. Building three or four rows yourself makes the add-the-two-above rule permanent in a way a formula sheet never does — it is exactly how a Bhanzu trainer would open the topic.
Conclusion
Pascal's Triangle is a number array where each interior entry is the sum of the two directly above it, and the edges are always $1$.
Row $n$ (apex is row $0$) 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, the triangular numbers on a diagonal, and Fibonacci numbers on shallow diagonals.
The most common mistakes are counting rows from $1$, dropping a variable in the expansion, and forgetting the $(n-r)!$ in the choose formula.
The triangle is the natural on-ramp to the binomial theorem and to combination counting in probability.
Practice These to Solidify Your Understanding
Work through these in order, then check each against the triangle:
Write out rows $0$ through $6$ of Pascal's Triangle.
Expand $(x + 1)^5$ using the correct row.
Compute $\binom{8}{3}$ with the choose formula and confirm it against row $8$.
In how many ways can exactly four heads appear in six coin tosses?
If the expansion in question $2$ goes wrong, come back to the "descending and ascending powers" mistake above. To build this with a live trainer who shows you the triangle before the formula, explore Bhanzu's algebra tutor or help with algebra sessions, or browse math classes online. Want a trainer to walk through more Pascal's Triangle problems? Book a free demo class.
Read More
Binomial — what a two-term expression is and how it behaves.
Factoring Trinomials — reversing a binomial expansion back into factors.
Cube of a Binomial — the $(a+b)^3$ expansion the triangle's row $3$ produces.
Multiplying Binomials — the long-hand method the triangle lets you skip.
Perfect Square Trinomial — the $(a+b)^2$ pattern that is row $2$ in disguise.
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