Previously, we studied combinations. In the shortcut to finding [latex]{\left(x+y\right)}^{n}[/latex], we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. In this case, we use the notation [latex]\left(\begin{gathered}n\\ r\end{gathered}\right)[/latex] instead of [latex]C\left(n,r\right)[/latex], but it can be calculated in the same way. So

[latex]\left(\begin{gathered}n\\ r\end{gathered}\right)=C\left(n,r\right)=\dfrac{n!}{r!\left(n-r\right)!}[/latex]

The combination [latex]\left(\begin{gathered}n\\ r\end{gathered}\right)[/latex] is called a binomial coefficient. An example of a binomial coefficient is [latex]\left(\begin{gathered}5\\ 2\end{gathered}\right)=C\left(5,2\right)=10[/latex].

When we expand [latex]{\left(x+y\right)}^{n}[/latex] by multiplying, the result is called a binomial expansion, and it includes binomial coefficients. If we wanted to expand [latex]{\left(x+y\right)}^{52}[/latex], we might multiply [latex]\left(x+y\right)[/latex] by itself fifty-two times. This could take hours! If we examine some simple binomial expansions, we can find patterns that will lead us to a shortcut for finding more complicated binomial expansions.

[latex]\begin{align}&{\left(x+y\right)}^{2}={x}^{2}+2xy+{y}^{2} \\ &{\left(x+y\right)}^{3}={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3} \\ &{\left(x+y\right)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4} \end{align}[/latex]

First, let’s examine the exponents. With each successive term, the exponent for [latex]x[/latex] decreases and the exponent for [latex]y[/latex] increases. The sum of the two exponents is [latex]n[/latex] for each term.

Next, let’s examine the coefficients. Notice that the coefficients increase and then decrease in a symmetrical pattern. The coefficients follow a pattern:

[latex]\left(\begin{gathered}n\\ 0\end{gathered}\right),\left(\begin{gathered}n\\ 1\end{gathered}\right),\left(\begin{gathered}n\\ 2\end{gathered}\right),\dots,\left(\begin{gathered}n\\ n\end{gathered}\right)[/latex].

These patterns lead us to the Binomial Theorem, which can be used to expand any binomial.

[latex]\begin{align}{\left(x+y\right)}^{n}& =\sum\limits _{k=0}^{n}\left(\begin{gathered}n\\ k\end{gathered}\right){x}^{n-k}{y}^{k} \\ & ={x}^{n}+\left(\begin{gathered}n\\ 1\end{gathered}\right){x}^{n - 1}y+\left(\begin{gathered}{c}n\\ 2\end{gathered}\right){x}^{n - 2}{y}^{2}+\dots+\left(\begin{gathered}n\\ n - 1\end{gathered}\right)x{y}^{n - 1}+{y}^{n} \end{align}[/latex]

Another way to see the coefficients is to examine the expansion of a binomial in general form, [latex]x+y[/latex], to successive powers 1, 2, 3, and 4.

[latex]\begin{align}&{\left(x+y\right)}^{1}=x+y \\ &{\left(x+y\right)}^{2}={x}^{2}+2xy+{y}^{2} \\ &{\left(x+y\right)}^{3}={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3} \\ &{\left(x+y\right)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4} \end{align}[/latex]

Can you guess the next expansion for the binomial [latex]{\left(x+y\right)}^{5}?[/latex]

Above is an illustration the following:

• There are [latex]n+1[/latex] terms in the expansion of [latex]{\left(x+y\right)}^{n}[/latex].
• The degree (or sum of the exponents) for each term is [latex]n[/latex].
• The powers on [latex]x[/latex] begin with [latex]n[/latex] and decrease to 0.
• The powers on [latex]y[/latex] begin with 0 and increase to [latex]n[/latex].
• The coefficients are symmetric.

To determine the expansion on [latex]{\left(x+y\right)}^{5}[/latex], we see [latex]n=5[/latex], thus, there will be 5+1 = 6 terms. Each term has a combined degree of 5. In descending order for powers of [latex]x[/latex], the pattern is as follows:

• Introduce [latex]{x}^{5}[/latex], and then for each successive term reduce the exponent on [latex]x[/latex] by 1 until [latex]{x}^{0}=1[/latex] is reached.
• Introduce [latex]{y}^{0}=1[/latex], and then increase the exponent on [latex]y[/latex] by 1 until [latex]{y}^{5}[/latex] is reached.
  [latex]{x}^{5},{x}^{4}y,{x}^{3}{y}^{2},{x}^{2}{y}^{3},x{y}^{4},{y}^{5}[/latex]

The next expansion would be

[latex]{\left(x+y\right)}^{5}={x}^{5}+5{x}^{4}y+10{x}^{3}{y}^{2}+10{x}^{2}{y}^{3}+5x{y}^{4}+{y}^{5}[/latex].

But where do those coefficients come from? The binomial coefficients are symmetric. We can see these coefficients in an array known as Pascal’s Triangle.

To generate Pascal’s Triangle, we start by writing a 1. In the row below, row 2, we write two 1’s. In the 3^rd row, flank the ends of the rows with 1’s, and add [latex]1+1[/latex] to find the middle number, 2. In the [latex]n\text{th}[/latex] row, flank the ends of the row with 1’s. Each element in the triangle is the sum of the two elements immediately above it.

To see the connection between Pascal’s Triangle and binomial coefficients, let us revisit the expansion of the binomials in general form.

Using the Binomial Theorem to Find a Single Term

Expanding a binomial with a high exponent such as [latex]{\left(x+2y\right)}^{16}[/latex] can be a lengthy process.

Sometimes we are interested only in a certain term of a binomial expansion. We do not need to fully expand a binomial to find a single specific term.

Note the pattern of coefficients in the expansion of [latex]{\left(x+y\right)}^{5}[/latex].

[latex]{\left(x+y\right)}^{5}={x}^{5}+\left(\begin{gathered}5\\ 1\end{gathered}\right){x}^{4}y+\left(\begin{gathered}5\\ 2\end{gathered}\right){x}^{3}{y}^{2}+\left(\begin{gathered}5\\ 3\end{gathered}\right){x}^{2}{y}^{3}+\left(\begin{gathered}5\\ 4\end{gathered}\right)x{y}^{4}+{y}^{5}[/latex]

The second term is [latex]\left(\begin{gathered}5\\ 1\end{gathered}\right){x}^{4}y[/latex]. The third term is [latex]\left(\begin{gathered}5\\ 2\end{gathered}\right){x}^{3}{y}^{2}[/latex]. We can generalize this result.

[latex]\left(\begin{gathered}n\\ r\end{gathered}\right){x}^{n-r}{y}^{r}[/latex]