Pascal's Triangle

The Central Idea

Pascal's triangle begins with a single \(1\):

Each new row begins and ends with \(1\).

Every interior entry is found by adding the two entries above it.

The first several rows are

These are not arbitrary numbers. They are binomial coefficients.

Rows and Binomial Coefficients

If the top row is called row \(0\), then row \(n\) contains

For example, row \(4\) is

Evaluating gives

Row Numbering

We will number the top row as row \(0\).

Therefore:

• row \(0\): \(1\);
• row \(1\): \(1,\ 1\);
• row \(2\): \(1,\ 2,\ 1\);
• row \(3\): \(1,\ 3,\ 3,\ 1\);
• row \(4\): \(1,\ 4,\ 6,\ 4,\ 1\).

This numbering matches the top index in \(\binom{n}{r}\).

Constructing the Triangle

Start with

Place \(1\)'s at the ends of every new row.

To find an interior entry, add the two numbers above it.

For example, from

the next row has interior entries

So the next row is

Why the Interior-Addition Rule Works

The recursive rule comes from Pascal's identity:

The two entries directly above \(\binom{n}{r}\) are precisely

and

Their sum creates the new entry.

A Counting Explanation of Pascal's Identity

Suppose we want to choose \(r\) objects from \(n\) objects and focus on one particular object \(A\).

Every selection falls into exactly one of two cases.

If \(A\) is excluded, choose all \(r\) objects from the remaining \(n-1\):

If \(A\) is included, choose the remaining \(r-1\) objects from the other \(n-1\):

Add the disjoint cases:

The First Eight Rows

Each row can be generated either recursively or directly from binomial coefficients.

Row \(5\)

Row \(5\) contains

Therefore,

Symmetry

Every row of Pascal's triangle is symmetric.

This follows from the identity

For example, row \(6\) is

Reading left to right or right to left gives the same sequence.

Why the Rows Are Symmetric

Choosing \(r\) objects from \(n\) determines exactly which \(n-r\) objects were not chosen.

Therefore, selecting the included objects and selecting the excluded objects are equivalent descriptions of the same division.

That gives

The Outer Diagonals

The outermost entries are all \(1\):

and

There is exactly one way to choose no objects and exactly one way to choose all objects.

The Next Diagonal

Moving inward one diagonal gives

because

Thus the natural numbers appear directly in Pascal's triangle.

The Triangular-Number Diagonal

The next diagonal contains

These are the triangular numbers.

They are given by

For example,

The Tetrahedral-Number Diagonal

The next diagonal contains

These values are

They are sometimes called tetrahedral numbers because of their geometric interpretation in three-dimensional figurate-number patterns.

Row Sums

The sum of row \(n\) is

In symbols,

Example: Sum of Row \(4\)

Row \(4\) is

Its sum is

Since

the row-sum identity holds.

Why Row Sums Are Powers of Two

An \(n\)-element set has

subsets of size \(r\).

Adding across all possible subset sizes counts every subset:

But every element independently has two choices:

• included;
• excluded.

Therefore, there are \(2^n\) subsets.

Alternating Row Sums

For every \(n\ge1\),

This means

Example: Alternating Sum of Row \(5\)

Row \(5\) is

Alternating signs gives

Therefore,

Even and Odd Subset Counts

The alternating-sum identity implies that for \(n\ge1\), an \(n\)-element set has equally many even-sized and odd-sized subsets.

Since there are \(2^n\) subsets total, each category contains

subsets.

The Hockey-Stick Pattern

Certain diagonal sums form another binomial coefficient.

The identity is

The shape traced through Pascal's triangle resembles a hockey stick.

Example: Hockey-Stick Sum

The values are

And

Why the Hockey-Stick Identity Works

Suppose we choose \(r+1\) objects from

Classify the selections by their largest selected element.

If the largest element is \(k+1\), choose the remaining \(r\) elements from the first \(k\) elements:

Adding over all possible largest elements produces the hockey-stick sum.

Shallow Diagonals and Fibonacci Numbers

Another striking pattern appears by summing entries along shallow diagonals.

The sums begin

which are Fibonacci numbers.

One version of the identity is

Example: A Fibonacci Diagonal Sum

For \(n=5\),

Therefore,

Indeed,

Powers of Eleven Pattern

The first few rows of Pascal's triangle resemble the digits of powers of \(11\):

This works directly only while every entry is a single digit.

Why Carrying Becomes Necessary

Row \(5\) is

These cannot simply be concatenated as decimal digits because \(10\) is not a single digit.

However, if decimal carrying is handled correctly, the relationship with powers of \(11\) continues.

This is an interesting numeral pattern, not the defining structure of Pascal's triangle.

Pascal's Triangle and Binomial Expansions

Row \(n\) gives the coefficients in the expansion of

For example, row \(4\) is

Therefore,

The Exponent Pattern in a Binomial Expansion

In the expansion of

the exponent of \(a\) decreases from \(n\) to \(0\), while the exponent of \(b\) increases from \(0\) to \(n\).

The general term is

Pascal's triangle supplies the coefficient \(\binom{n}{r}\).

Example: Expand \((x+y)^5\)

Row \(5\) is

Therefore,

Why Binomial Coefficients Appear in the Expansion

Write

To produce a term containing exactly \(r\) copies of \(b\), choose which \(r\) of the \(n\) factors contribute \(b\).

There are

ways to make that choice.

Therefore, the coefficient of \(a^{n-r}b^r\) is \(\binom{n}{r}\).

Pascal's Triangle as a Recursive Algorithm

Pascal's triangle provides a simple algorithm for computing binomial coefficients without evaluating factorials.

1. Start with \(1\).
2. Place \(1\)'s at the ends of each new row.
3. Add adjacent entries from the previous row to create each interior entry.
4. Repeat until the desired row is reached.

Example: Build Row \(6\) from Row \(5\)

Start with row \(5\):

Add adjacent pairs:

Add \(1\)'s at the ends:

Computing a Single Entry Recursively

To find

Pascal's identity gives

Since

and

we obtain

Pascal's Triangle and Lattice Paths

Binomial coefficients also count shortest lattice paths.

To move \(r\) steps right and \(u\) steps up, choose which

of the

total step positions are rightward.

Therefore, the number of shortest paths is

Why Lattice Paths Follow Pascal's Recursion

Any path reaching a given lattice point must arrive from one of two neighboring points:

• from the left;
• from below.

Therefore, the number of paths to the point is the sum of the path counts to those two predecessor points.

This is exactly the same addition rule used in Pascal's triangle.

Pascal's Triangle as a Dynamic Program

The recursive construction is an example of a broader computational idea: build larger solutions from previously computed smaller solutions.

Instead of recalculating

from factorials each time, we reuse nearby values.

This is an elementary example of dynamic programming.

Local Rule, Global Structure

Pascal's triangle is generated by a very simple local rule:

Yet this local rule produces global patterns involving:

• symmetry;
• powers of two;
• figurate numbers;
• Fibonacci numbers;
• binomial expansions;
• subset counts;
• lattice paths.

This is a recurring theme in discrete mathematics: simple recursive rules can generate rich large-scale structure.

Parity Patterns

If we mark every odd entry and leave every even entry unmarked, Pascal's triangle develops a repeating triangular pattern.

At larger scales, this pattern is closely related to the Sierpiński triangle.

This occurs because binomial coefficients have strong modular patterns, especially modulo \(2\).

A full treatment belongs to later discrete and number-theoretic topics, but the pattern is worth recognizing here.

Entries Modulo Two

Reducing each entry modulo \(2\) changes every even number to \(0\) and every odd number to \(1\).

For example, row \(4\)

becomes

Row \(5\)

becomes

Pascal's Triangle Is Infinite

The triangle has no final row.

For every nonnegative integer \(n\), there is a row containing

entries:

The displayed rows are only the beginning of an infinite recursive array.

Each Row Has \(n+1\) Entries

Row \(n\) runs from

through

Therefore, row \(n\) contains

entries.

The Middle of a Row

Binomial coefficients increase toward the center of each row and then decrease symmetrically.

If \(n\) is even, the unique largest central entry is

If \(n\) is odd, the two central entries are equal:

Example: Middle of Row \(8\)

Row \(8\) is

The largest entry is the center:

Example: Middle of Row \(7\)

Row \(7\) is

The two middle entries are

Pascal's Triangle and Probability

Binomial coefficients count the possible locations of successes among repeated trials.

For example, in \(n\) trials, exactly \(r\) successes can occur in

positional patterns.

This is why Pascal's triangle appears naturally in binomial probability calculations.

Example: Four Trials

With four success-failure trials, the possible counts by number of successes are

Thus the counts are

These are exactly row \(4\) of Pascal's triangle.

Pascal's Triangle and Repeated-Object Arrangements

The coefficient

also counts distinguishable arrangements of \(r\) identical objects of one type and \(n-r\) identical objects of another type.

The repeated-object formula gives

Thus every entry in Pascal's triangle has a repeated-arrangement interpretation.

Pascal's Triangle and Binary Strings

The entry

counts binary strings of length \(n\) containing exactly \(r\) ones.

Therefore, row \(n\) classifies all \(2^n\) binary strings according to their number of ones.

Example: Binary Strings of Length Five

Row \(5\) is

Therefore:

• one string has zero \(1\)'s;
• five strings have one \(1\);
• ten strings have two \(1\)'s;
• ten strings have three \(1\)'s;
• five strings have four \(1\)'s;
• one string has five \(1\)'s.

The total is

Common Error: Starting Row Numbers at One

Different sources sometimes number rows differently.

In this lesson, the top \(1\) is row \(0\).

This convention ensures row \(n\) contains

Always check the row-numbering convention before interpreting an exercise.

Common Error: Adding the Wrong Pair

Each interior entry is the sum of the two entries immediately above it:

Do not add entries from the same row or skip positions.

Common Error: Forgetting the Edge Ones

Every row begins and ends with

These boundary entries correspond to

and

Common Error: Treating the Triangle as a Memorized Number Pattern

Pascal's triangle should not be treated as a list of unrelated rows to memorize.

Its structure comes from

and Pascal's identity.

Understanding those principles allows any row to be reconstructed.

Common Error: Using the Powers-of-Eleven Pattern as the Definition

The resemblance to powers of \(11\) is a secondary numeral pattern.

It becomes inconvenient once row entries exceed one digit.

Pascal's triangle is fundamentally defined by binomial coefficients and the recursive addition rule.

Common Error: Mixing Row and Entry Indices

In

\(n\) identifies the row.

The value \(r\) identifies the position within the row, beginning with \(r=0\).

Thus the fourth entry of row \(6\), counting from \(r=0\), is

A Reliable Procedure for Building Pascal's Triangle

1. Begin with row \(0\): \(1\).
2. Place \(1\) at both ends of the next row.
3. Add adjacent entries from the previous row.
4. Place each sum between its parent entries.
5. Repeat until the required row is reached.
6. Check that the row is symmetric.
7. Check that row \(n\) has \(n+1\) entries.
8. Check that the row sum equals \(2^n\).

A Reliable Procedure for Reading an Entry

1. Identify the row number \(n\).
2. Count positions from \(r=0\).
3. Write the entry as \(\binom{n}{r}\).
4. Use symmetry if the mirrored position is easier.
5. Evaluate by factorials, recursion, or known rows as appropriate.

Example: Full Construction Workflow

Construct row \(7\) from row \(6\).

Row \(6\) is

Add adjacent pairs:

Add boundary \(1\)'s:

Check the row sum:

Practice 1: Build the Next Row

Starting from

construct the next row.

Practice 2: Identify an Entry

What is the value of

Practice 3: Row Sum

What is the sum of all entries in row \(9\)?

Practice 4: Symmetry

Use symmetry to evaluate

Practice 5: Pascal's Identity

Evaluate

using

and

Practice 6: Hockey-Stick Pattern

Evaluate

Practice 7: Binary Strings

How many binary strings of length \(8\) contain exactly three \(1\)'s?

Practice 8: Binomial Expansion

Use Pascal's triangle to expand

Practice 9: Alternating Sum

Evaluate

Practice 10: Identify the Error

A student says the row after

is

Identify the incorrect entry and construct the correct row.

Practice Answers

Practice 1

Add adjacent entries and place \(1\)'s at the ends:

Practice 2

Practice 3

Row \(9\) sums to

Practice 4

Practice 5

Pascal's identity gives

Therefore,

Practice 6

By the hockey-stick identity,

Therefore,

Practice 7

Choose which three positions contain \(1\)'s:

Practice 8

Row \(4\) is

Therefore,

Practice 9

This is the alternating sum of row \(6\):

Practice 10

The middle-left interior entry should be

not \(11\).

The correct next row is

What to Remember

Row \(n\) of Pascal's triangle contains

Every row begins and ends with \(1\).

Every interior entry satisfies

Every row is symmetric:

The sum of row \(n\) is

Pascal's triangle also exposes diagonal figurate numbers, hockey-stick sums, Fibonacci patterns, binary-string counts, lattice paths, probability counts, and the coefficients of binomial expansions.

The triangle is best understood not as a memorized array but as a visual organization of binomial coefficients and their recursive relationships.

Where We Go Next

Pascal's triangle provides the coefficients of powers of a binomial, but the relationship extends beyond the triangle itself into general algebraic expansion formulas.

The next lesson, Binomial and Multinomial Theorems, develops those formulas systematically and shows how combination and multinomial coefficients encode the terms in polynomial expansions.