# What Is Pascal’s Triangle?

Pascal’s triangle is an amazing tool, especially when you need to calculate expressions on the form ${\left(a+b\right)}^{n}$. There are two ways to represent Pascal’s triangle: Through binomial coefficients or through finding the next row by adding together the neighbors of the row above.

Theory

### Method1

Pascal’s triangle is built up of the binomial coefficients $\left(\genfrac{}{}{0.0pt}{}{n}{k}\right)$, where $n$ is the row and $k$ is the place in the row.

You need to remember that Pascal’s triangle both start the row number and place number with 0. When you insert the numbers for $n$ and $k$ into the binomial coefficients, Pascal’s triangle is formed. It looks like this:

Using this method makes it easy to find the exact coefficient you need.

Note! In general, you have that $\begin{array}{llll}\hfill \left(\genfrac{}{}{0.0pt}{}{n}{0}\right)& =1=\left(\genfrac{}{}{0.0pt}{}{n}{n}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \left(\genfrac{}{}{0.0pt}{}{n}{1}\right)& =n=\left(\genfrac{}{}{0.0pt}{}{n}{n-1}\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Convince yourself that this is correct!

Example 1

Find the numbers in row number 3 in Pascal’s triangle

Looking at row number 3 means that $n=3$ and $k$ is 0, 1, 2 and 3. That makes the numbers in row number 3 as follows:

If you calculate the values of the binomial coefficients, you get this triangle:

Note! Every row of Pascal’s triangle starts and ends with 1.

But if you can’t find or use the binomial coefficients, what should you do? There is a very simple method you can use if you already have the above row, and it goes like this:

Theory

### Method2

When you add together two numbers that are next to each other in a row, you get the number that goes between the two on the row below.

Example 2

Write down the first four rows of Pascal’s triangle by adding together row entries

First row

has one entry and is row number 0. The entry is 1.

Second row

has two entries and is row number 1. Both entries are 1, because all rows in Pascal’s triangle start and end with 1.

Third row

has three entries and is row number 2. Now it pays off to have written down the first two rows, because you’re going to find the third row through Method 2. You know the first and last entries have to be 1, so the only entries you need to fill is the one in the middle.

Here you can see that the middle entry of the third row is $1+1=2$.

Fourth row

has four entries and is row number 3. The first and last entry has to be 1, so the entries you need to find are the two in the middle of the row. This is what comes out:

Here you can see that $1+2=3$ and that $2+1=3$, which is how you get the two middle entries on the fourth row (row number 3).

You can always add a row to Pascal’s triangle by using one of the methods outlined above, but if you need a row far down the triangle it is easiest to use the binomial coefficient $\left(\genfrac{}{}{0.0pt}{}{n}{k}\right)$, where $n$ is the row and $k$ is the place in the row. Of course you can use Method 2 as well, but if you want row number 30 it’ll take a very long time.

Among other things, Pascal’s triangle is used to find the number of combinations, and calculations on the form ${\left(a+b\right)}^{n}$ (Newton’s binomial theorem).

Example 3

The figure below shows a part of Pascal’s triangle.

From what you can see in the figure, find the values of $x$ and $y$.

You know that the sum of two neighbors is equal to the number between them on the row below. From this you can find two equations with two unknowns: $\begin{array}{lll}\hfill 21+x& =y,\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\phantom{\rule{0.33em}{0ex}}\\ \hfill y+2x& =126.\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\phantom{\rule{0.33em}{0ex}}\end{array}$

Solve (1) for $x$:

 $x=y-21.$ (3)

Solve (2) for $y$ and insert (3): $\begin{array}{llllll}\hfill y+2x& =126\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =126-2x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =126-2\left(y-21\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =126-2y+42\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 3y& =168\phantom{\rule{2em}{0ex}}& \hfill & |:3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =56.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Finally, you find $x$ by inserting for $y$ in (3):

 $x=56-21=35.$

This means the missing numbers in Pascal’s triangle are 35 and 56.

Example 4

Here you’ll see how to use the binomial theorem to calculate expressions on the form ${\left(a+b\right)}^{n}$ for $n=0,1,2,3,4$:

$\begin{array}{llll}\hfill {\left(a+b\right)}^{0}& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {\left(a+b\right)}^{1}& =1\cdot a+1\cdot b\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {\left(a+b\right)}^{2}& =1\cdot {a}^{2}+2\cdot ab+1\cdot {b}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {\left(a+b\right)}^{3}& =1\cdot {a}^{3}+3\cdot {a}^{2}b+3\cdot a{b}^{2}+1\cdot {b}^{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {\left(a+b\right)}^{4}& =1\cdot {a}^{4}+4\cdot {a}^{3}b+6\cdot {a}^{2}{b}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}+4\cdot a{b}^{3}+1\cdot {b}^{4}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill {\left(a+b\right)}^{0}& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {\left(a+b\right)}^{1}& =1\cdot a+1\cdot b\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {\left(a+b\right)}^{2}& =1\cdot {a}^{2}+2\cdot ab+1\cdot {b}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {\left(a+b\right)}^{3}& =1\cdot {a}^{3}+3\cdot {a}^{2}b+3\cdot a{b}^{2}+1\cdot {b}^{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {\left(a+b\right)}^{4}& =1\cdot {a}^{4}+4\cdot {a}^{3}b+6\cdot {a}^{2}{b}^{2}+4\cdot a{b}^{3}+1\cdot {b}^{4}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

You can see that the coefficients are the same as the first four rows of Pascal’s triangle.

Example 5

Calculate ${\left(2x+1\right)}^{4}$

You could find this by multiplying out all the parentheses, but why bother when you could just use Pascal’s triangle like we showed in Example 4. The first thing to check is the exponent, which is 4. That tells you that row number 4 in Pascal’s triangle has the coefficients you need (remember that the first row is row number 0). This gives these the coefficients:

Finding the binomial coefficients gives you:

Next, you write out the multiplications of the two terms in ${\left(2x+1\right)}^{4}$. The method is as follows:

• The parenthesis consists of two terms, $2x$ and 1. These are being raised to a power by an exponent.

• The exponent of the first term, $2x$, needs to start with the exponent of the parenthesis, which in this case is 4, and decrease to 0 term by term.

• The exponent of the second term, 1, needs to start at 0 and increase to the exponent of the parenthesis, 4, term by term.

That looks like this: $\begin{array}{llll}\hfill {\left(2x\right)}^{4}\cdot {1}^{0}& =16{x}^{4}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {\left(2x\right)}^{3}\cdot {1}^{1}& =8{x}^{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {\left(2x\right)}^{2}\cdot {1}^{2}& =4{x}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {\left(2x\right)}^{1}\cdot {1}^{3}& =2x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {\left(2x\right)}^{0}\cdot {1}^{4}& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Now you can put the five coefficients in front of each of the five terms. That gives you this:

$\begin{array}{llll}\hfill {\left(2x+1\right)}^{4}& =1\cdot 16{x}^{4}+4\cdot 8{x}^{3}+6\cdot 4{x}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}+4\cdot 2x+1\cdot 1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =16{x}^{4}+32{x}^{3}+24{x}^{2}+8x+1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill {\left(2x+1\right)}^{4}& =1\cdot 16{x}^{4}+4\cdot 8{x}^{3}+6\cdot 4{x}^{2}+4\cdot 2x+1\cdot 1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =16{x}^{4}+32{x}^{3}+24{x}^{2}+8x+1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

When you get comfortable with the method, you’ll just execute the final calculation. Everything that comes before are just lines of reasoning going on in your head. This is much faster than actually calculating the parenthesis the old-fashioned way, so it’s smart to get used to using Pascal’s triangle!