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Pascal's Triangle

The math behind Pascal’s Triangle is very simple. It is so simple that even young students can learn and repeat the pattern.

But as you look more closely, you start to notice interesting patterns. These patterns lead to powerful ideas in counting and other areas of math.

Remember: While Pascal’s Triangle helps count combinations where order does not matter (like combinations), permutations are used when the order of the selected items matters.

Let’s start by learning how the arithmetic works.

The foundation of the triangle is $1$s. The first row is a $1.$ The second row is two $1$s. After that, the first and last entry in each row is a $1.$ Let's look at the first $3$ rows:

The question mark is the first entry that is not a $1.$ It is easily found by looking at the two entries it is between that are in the row above it. In this case the "?" is $1+1=2$. Here is a look at the first $4$ rows. Note: the general convention is for the top row to be referred to as row $0.$ Thus the first $4$ rows are rows $0$ through $3.$

As you can see, the next row is $1, 3, 3, 1.$ The $3$s come from $1+2=3$ and $2+1=3$.

The first and last terms are $1$s like every other row. The $4$ comes from the sum of the two numbers above it, $1+3=4$, the $6=3+3$, and then the second $4$ is $3+1=4$.

Another important thing to note about the Pascal's Triangle is the first entry in each row is always a $1$ and is called entry $0.$ So in row $4,$ entry $2$ is a $6.$ Row $0$ through row $8$ are shown below. Make sure you know where each number comes from before moving on...

Being able to generate new rows of the Pascal's Triangle is step $1$ of a long path that can lead to many interesting insights and wonders that such a simple pattern can be used to better understand a diverse set of mathematical situations. Lets begin by making some basic observations.

1. The first pattern is in red and is simply the counting numbers.

1. The second pattern is underlined and in blue. It is a special sequence called the trianglular numbers. The triangular numbers are the sum of the first $n$ counting numbers. $$1=1$$ $$1+2=3$$ $$1+2+3=6$$ $$1+2+3+4=10$$ $$1+2+3+4+5=15$$

1. A third important thing to notice is that each line is symmetric. This means it reads the same forwards and backwards.

1. Lets look at the sum of the entries in each row. $$1=1$$ $$1+1=2$$ $$1+2+1=4$$ $$1+3+3+1=8$$ $$1+4+6+4+1=16$$ $$1+5+10+10+5+1=32$$

See if you can notice the pattern.

The pattern is the sum of each row is the next power of 2. Row 0 is $2^0=1$, row 4 is $2^4=16$, and for the question, row 7 is $2^7=128$.

1. The Hockey Stick Principle - The hockey stick principle is a surprising relationship between diagonals of the Pascal's Triangle. Essentially, by looking at a diagonal starting with the $1$ and then drawing a hockey stick, you can quickly and easily find the sum of the diagonal. We will further discuss this later. The Pascal's Triangle below shows what is meant by the hockey stick principle. Note that $1+2+3+4+5+6=21$ and $1+4+10=15$. These are the two cases highlighted in the picture below.

The answer is $252$ as it is in row $10$ and diagonal from $126$ in row $9.$ Now hopefully you are a little skeptical of the value of the hockey stick principle. After all, it may be easier to add those $6$ numbers together rather than generate $11$ rows of the Pascal's Triangle. Until you learn how to generate specific entries in the triangle without having to generate the entire triangle. This is the next (and last) of our basic observations about the triangle. Next we will begin discussing the applications of the triangle.

1. It turns out that the entries in the Pascal's Triangle can be generated using the very important Combinations formula. This formula is used to determine how many ways to pick $r$ things from a group of $n$ things. Order does not matter and it can be generalized to imagine the selections happening simultaneously.

The formula is: $$nCr=\frac{n(n-1)(n-2)...(n-r+1)}{r!}$$

For example: $$10C4 = \frac{10(9)(8)(7)}{4(3)(2)(1)}=210$$

This is found by: $$15C6=\frac{15(14)(13)(12)(11)(10)}{(6(5)(4)(3)(2)(1)}$$ Which can be simplied to: $$7(13)(11)(5)=5005$$

1. Binomial Expansion - Binomial expansion is the process of multiplying out binomials raised to a certain power. Some basic binomial expansions are below. These can be verified by applying your favorite way to muliply binomials (FOIL, double distributive property, the smiley face method, etc.).

If you look closely at the coefficients of the variables, you should be able to spot the rows in the Pascal's Triangle. Row $4$ of the Pascal's Triangle is $1, 4, 6, 4, 1$, which are exactly the coefficients of each term of $(a+b)^4$. It is important to note two things from this. First, row number and the power are the same. This holds true for the connection between all binomial powers and the Pascal's Triangle. Second, the degree (sum of the powers in each term) of each term of the expansion is equal to the row number and power as well, which in this case 4. You can also count the entries based on the exponent of $b$. $a^4$ has a coefficient of $1$ which is entry $0$ of row $4$ because there is an implied ("invisible") $b^0$ factor. $a^3b^1$ has a coefficient of $4$ because it is entry $1$ in the $4$th row and $1$ is the exponent on $b$. We can further connect this to combinations like this:

$a^2b^2$ has a coefficient of $6$ $\longrightarrow$ $6$ is entry $2$ in row $4$ of the Pascal's Triange $\longrightarrow$ entry $2$ of row $4$ can be found by $4C2$ $\longrightarrow$ the coefficient of $a^2b^2$ can be found by $4C2$.

Solution: The $a^3b^9$ term has a degree of $12$ and $a$ has exponent $3,$ so the coefficient is $12C3$.

Think about the case of $(a+3)^6$. When it is not the expansion of binomial in two variables.

The point of tension comes from the second term in the binomial not being a variable. A nice way of keeping track of things is to look at it like this. Row 6 of the Pascal's Triangle is: $$1\hspace{1cm}6\hspace{1cm}15\hspace{1cm}20\hspace{1cm}15\hspace{1cm}6\hspace{1cm}1$$ Thus, the expansion is: $$1(a^6)+6(a^5)(3^1)+15(a^4)(3^2)+20(a^3)(3^3)+15(a^2)(3^4)+6(a)(3^5)+3^6,$$ which equals $$a^6+18a^5+135a^4+540a^3+1215a^2+1458a+729.$$

Pascal’s Triangle is a triangle made of numbers. As the name suggests, the numbers are arranged in rows that grow longer as you move down the triangle.

Each number in the triangle follows a simple rule, which makes the pattern easy to build and remember. Even though the rule is simple, it leads to many interesting results.

Pascal’s Triangle has many special properties that show up as you explore it more. These patterns are useful in areas like number theory, combinatorics, and counting and probability. The entries in Pascal’s Triangle are binomial coefficients. Combinatorics with Pascal’s Triangle shows how these values can also be interpreted as counting different selections.

Later on, we will look at some of these properties and see how they are used in math problems.

Let's take a look at the first few rows of the triangle: $$\begin{array}{c} 1 \\ 1\,\,\,\,1 \\ 1\,\,\,\,2\,\,\,\,1 \\ 1\,\,\,\,3\,\,\,\,3\,\,\,\,1 \\ 1\,\,\,\,4\,\,\,\,6\,\,\,\,4\,\,\,\,1 \\ 1\,\,\,\,5\,\,\,10\,\,10\,\,\,5\,\,\,\,1 \\ \cdots \end{array}$$

Let's think about where these numbers come from.

With the exception of the $1$ at the top of the triangle, each number is the sum of the two numbers above it. Let's look at the above rows again, this time with some numbers colored for reference: $$\begin{array}{c} {\color{#8925AE}{1}} \\ {\color{#8925AE}{1}}\,\,\,\,1 \\ 1\,\,\,\,{\color{blue}{2}}\,\,\,\,{\color{blue}{1}} \\ 1\,\,\,\,3\,\,\,\,{\color{blue}{3}}\,\,\,\,1 \\ 1\,\,\,\,{\color{green}{4}}\,\,\,\,{\color{green}{6}}\,\,\,\,4\,\,\,\,1 \\ 1\,\,\,\,5\,\,\,{\color{green}{10}}\,\,10\,\,\,5\,\,\,\,1 \\ \cdots \end{array}$$

• ${\color{#8925AE}{0+1 = 1}}$. Note that $1$s populate the far left and right sides of Pascal's triangle.
• ${\color{blue}{2+1 = 3}}$.
• ${\color{green}{4+6 = 10}}$.

Note that the way Pascal's triangle is constructed leads to an inherent symmetry. Any given row of the triangle is the same forwards and backwards.

Now, let's start looking at some cool properties of Pascal's triangle! First, if we sum all the numbers in each row, we get the following: $$\begin{array}{c c} \text{Row} & \text{Sum} \\ \hline 1 & 1 = 2^0\\ 1\,\,\,\,1 & 2 = 2^1\\ 1\,\,\,\,2\,\,\,\,1 & 4 = 2^2\\ 1\,\,\,\,3\,\,\,\,3\,\,\,\,1 & 8 = 2^3\\ 1\,\,\,\,4\,\,\,\,6\,\,\,\,4\,\,\,\,1 & 16 = 2^4\\ 1\,\,\,\,5\,\,\,10\,\,10\,\,\,5\,\,\,\,1 & 32 = 2^5\\ \cdots \end{array}$$ It turns out that each row sums to a power of $2$. Each number in the triangle is counted towards two numbers in the next row. For example, note the highlighted numbers below: $$\begin{array}{c} 1 \\ 1\,\,\,\,1 \\ 1\,\,\,\,2\,\,\,\,1 \\ 1\,\,\,\,3\,\,\,\,{\color{#8925AE}{3}}\,\,\,\,1 \\ 1\,\,\,\,4\,\,\,\,{\color{#8925AE}{6}}\,\,\,\,{\color{#8925AE}{4}}\,\,\,\,1 \\ 1\,\,\,\,5\,\,\,10\,\,10\,\,\,5\,\,\,\,1 \\ \cdots \end{array}$$ The ${\color{#8925AE}{3}}$ counts towards both $3+{\color{#8925AE}{3}} = {\color{#8925AE}{6}}$ and ${\color{#8925AE}{3}}+1 = {\color{#8925AE}{4}}$.

Another important thing to know about Pascal's triangle is that the rows are zero-indexed, i.e. the first row of the triangle is denoted as row $0$. This implies that row $n$ contains $(n+1)$ elements. Looking at the sums of each row, this actually makes a lot of sense! The nth row of Pascal's triangle always sums to $2^n$.

We can also find some cool patterns in the diagonals of Pascal's triangle! $$\begin{array}{c} 1 \\ 1\,\,\,\,{\color{#8925AE}{1}} \\ 1\,\,\,\,{\color{#8925AE}{2}}\,\,\,\,{\color{blue}{1}} \\ 1\,\,\,\,{\color{#8925AE}{3}}\,\,\,\,{\color{blue}{3}}\,\,\,\,1 \\ 1\,\,\,\,{\color{#8925AE}{4}}\,\,\,\,{\color{blue}{6}}\,\,\,\,4\,\,\,\,1 \\ 1\,\,\,\,{\color{#8925AE}{5}}\,\,\,{\color{blue}{10}}\,\,10\,\,\,5\,\,\,\,1 \\ 1\,\,\,\,{\color{#8925AE}{6}}\,\,\,{\color{blue}{15}}\,\,\,20\,\,\,15\,\,\,6\,\,\,\,1 \\ \cdots \end{array}$$ The red numbers form the sequence of natural numbers $$1,2,3,4,5,6,\dots$$ whereas the blue numbers form the triangular numbers $$1,3,6,10,15,\dots$$ The reason the triangular numbers are named that way is because we can rewrite them as $$(1),(1+2),(1+2+3),(1+2+3+4),(1+2+3+4+5),\dots$$

Congratulations on finishing this explanation! As mentioned at the start, Pascal's triangle has many different applications.

• We can rewrite Pascal's triangle in terms of combinations: $$\begin{array}{c} \binom{0}{0} \\ \binom{1}{0}\,\,\,\,\binom{1}{1} \\ \binom{2}{0}\,\,\,\,\binom{2}{1}\,\,\,\,\binom{2}{2} \\ \binom{3}{0}\,\,\,\,\binom{3}{1}\,\,\,\,\binom{3}{2}\,\,\,\,\binom{3}{3} \\ \binom{4}{0}\,\,\,\,\binom{4}{1}\,\,\,\,\binom{4}{2}\,\,\,\,\binom{4}{3}\,\,\,\,\binom{4}{4} \\ \cdots \end{array}$$
• As a consequence of the above, Pascal's triangle is very useful for binomial expansion.
• There is a way to construct the Fibonacci sequence from Pascal's triangle! Hint: try summing some strategically placed diagonals.
• Compare the powers of $11$ to the rows of Pascal's triangle. Hints: $11 = 10 + 1$. Also, binomial expansion is very useful.

NOTE: read the other explanations to find out what a Binomial Coefficient is.

In this explanation we will be studying something called Pascal's Triangle, and how it relates to binomial coefficients. It is essentially a triangle of numbers that look like this:

Note that the numbers on the left and right edge of the triangle are always $1$. In addition, each number is generated by adding the two numbers to the top-left and top-right. For example, the $2$ is generated by adding the $1$ and the $1$ directly to its top-left and right. Similarly, the $3$ is created by adding the $2$ and the $1$.

But what can this triangle possibly have to do with binomial coefficients?

You may have noticed in the third row that $\binom{4}{0}=1$, $\binom{4}{1}=4$, $\binom{4}{2}=6$, $\binom{4}{3}=4$, and $\binom{4}{4}=1$. Following this logic, the entire triangle can be rewritten like this:

You can confirm that each of the values of the binomials equals to the corresponding value of the number on the triangle. We can see that the top number of each binomial also corresponds to the row number of pascals triangle (WARNING: the very top row with one number is considered row $0$) and the bottom number of each binomial is the element number in each row (element 0 being the first number in each row).

But how can this be so? Is this all a big coincidence, or is there an underlying reason?

There are several ways to see that this is indeed not a coincidence. The argument I will use is an algebraic argument.

Note that in Pascal Triangle, we have that the sum of two adjacent numbers forms the number that is directly below and adjacent to the two numbers.

So if the first number is $\binom{n}{k}$ and the second is $\binom{n}{k+1}$, then the number that they should sum to is $\binom{n+1}{k+1}$.

Recalls that the algebraic definition of a binomial is $\binom{n}{k}=\dfrac{n!}{k!(n-k)!}$.

Thus, we want to prove that $$\dfrac{n!}{k!(n-k)!}+\dfrac{n!}{(k+1)!(n-k-1)!}=\dfrac{(n+1)!}{(k+1)!(n-k)!}$$

Setting a common denominator to the first two fractions, we see that $$\begin{align*}\dfrac{n!}{k!(n-k)!}+\dfrac{n!}{(k+1)!(n-k-1)!}&=\dfrac{(k+1)n!}{(k+1)!(n-k)!}+\dfrac{(n-k)n!}{(k+1)!(n-k)!}\\ &= \dfrac{(n+1)n!}{(k+1)!(n-k)!}\\ &=\dfrac{(n+1)!}{(k+1)!(n-k)!}\end{align*}$$

Thus, we have proved that $\binom{n}{k}+\binom{n}{n+1}=\binom{n+1}{k+1}$, and this proves that the relationship between Pascal's Triangle and binomial coefficients was not a coincidence.

This also brings to light a property of binomial coefficients, known as Pascal's Identity: $$\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1}$$

Let's start with an example: $$(x+y)^2 = x^2 + 2xy + y^2$$ The coefficients ($1$, $2$, and $1$) happen to be the same as the second row of Pascal's Triangle, and when we try $$(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3,$$ we get the third row of Pascal's Triangle ($1$ $3$ $3$ $1$) as coefficients!

And as you may know, you can find any number in Pascal's Triangle by calculating: $\tbinom na$; where $n$ indicates the row and $a$ the place on the row.

Ex: $\tbinom 42 = 6$ tells us that the second place on the fourth row of Pascal's Triangle is a six.

We just saw that the coefficients of $(x+y)^3$ were the numbers of the third row. And that you calculate these coefficients by doing: $\tbinom 3a$, where a is the place on the row. So $$(x+y)^n = \tbinom n0 x^n + \tbinom n1 x^{n-1}y + \tbinom n2 x^{n-2}y^2 ... \tbinom nn y^n$$ (Note that $\tbinom 73 = \tbinom 74$.)

The proof and more about the binomial theorem can be found here.

Pascal's Triangle, named after Blaise Pascal, is a triangle where two numbers added up, result in the next number:

The top row of the triangle, containing only a single $1$, is indexed as row $0$. The next row of the triangle, containing two $1$s, is therefore row $1$.

Any of the numbers can be calculated via the expression $$\binom na$$ where $n$ is the row, and $a$ the place on that row. For example, $$\binom 43 = 6.$$

This also means that $$\binom 00 = 1.$$

(Video) Binomial Expansion Using Pascal's Triangle

by Mathispower4u

This video by Mathispower4u demonstrates how to expand using Pascal's Triangle and the binomial theorem.

Background

Before we discuss what exactly Pascal's triangle is, we briefly discuss what factorials and binomial coefficients are, because those are important for understanding why we care about Pascal's triangle.

A factorial of a nonnegative integer $n$ is the number of ways to arrange $n$ objects in a line, and is denoted as $n!$. We say that $0! = 1$ because there is one way to arrange zero objects, and that "way" is to do absolutely nothing. For positive integers this definition is more intuitive.

For instance, we can arrange the three letters $A, B,$ and $C$ in $6$ different ways:

This is precisely equal to

Now we define a binomial coefficient of two nonnegative integers $n$ and $k$ (where $n \geq k$) to be

You may recognize this formula from our lesson of combinations. This combinations formula tells us the number of combinations, which helps with binomial expansion.

We can plug numbers into the formula just like any other formula, but it helps to understand where the formula comes from. The quantity ${{n}\choose{k}}$ represents the number of ways to select $k$ objects out of a group of $n.$ This activity doesn't make physical sense if $k > n,$ so we just worry about when $n \geq k.$ We also assume natural numbers.

Ok, now that we've had a primer on binomial coefficients we can move to Pascal's triangle. Pascal's triangle is literally an upward-pointing triangle of Binomial coefficients. It looks like this:

The catch is that it goes downward forever, with every row having one more number than the previous. We can also evaluate all of the binomial coefficients to just rewrite Pascal's triangle with whole numbers:

We do not write the numbers in the triangle to have perfectly aligned columns, and the reason for this is simple: any number in the triangle is equal to the two numbers directly above it (except, of course, the numbers on the edges of the triangle, which are all equal to $1$). This phenomenon can be translated into a formula known as Pascal's Identity:

Put another way, if we add together any two adjacent numbers on the same row, the result equals the number directly beneath the ones we added together. If we use this formula enough times, we can compute any number on Pascal's Triangle.