Recall that the binomial theorem is an algebraic method of expanding a binomial that is raised to a certain power, such as $(4x+y)^7$. The theorem is given by the formula: 

$\displaystyle { (x+y) }^{ n }=\sum _{ k=0 }^{ n }{ \begin{pmatrix} n \\ k \end{pmatrix} } { x }^{ n-k }{ y }^{ k }=\sum _{ k=0 }^{ n }{ \begin{pmatrix} n \\ k \end{pmatrix} } { x }^{ k }{ y }^{ n-k }$

The coefficients that appear in the binomial expansion are called binomial coefficients. These are usually written $\displaystyle{\begin{pmatrix} n \\ k \end{pmatrix}}$or $_{n}{C}_{k}$, and pronounced "$n$ choose $k$".

The coefficient of a term $x^{n−k}y^k$ in a binomial expansion can be calculated using the combination formula. Recall that the combination formula represents the number of ways to choose $k$ objects from among $n$, where order does not matter. The formula consists of factorials:

$\displaystyle \begin{pmatrix} n \\ k \end{pmatrix} = \frac { n! }{ k!(n-k)! }$

Note that although this formula involves a fraction, the binomial coefficient $\begin{pmatrix} n \\ k \end{pmatrix}$ is actually an integer.

In calculating coefficients, recall that the factorial of a non-negative integer $n$, denoted by $n!$, is the product of all positive integers less than or equal to $n$. For example, $5!=5×4×3×2×1=120$. The value of $0!$ is $$$1$, according to the convention for an empty product. 

Finally, you may recall that the factorial $n!$ also represents the number of possible permutations of $n$, or the number of ways in which $n$ objects can be arranged or selected.

Example: Use the binomial formula to find the expansion of $(x+y)^4$

Start by substituting $n=4$ into the binomial formula:

$\displaystyle (x+y)^4=\sum _{ k=0 }^{ 4 }{ \begin{pmatrix} 4 \\ k \end{pmatrix} } { x }^{ 4-k }{ y }^{ k }$

In order to solve this, we will need to expand the summation for all values of $k$.

$\displaystyle \begin{aligned} (x+y)^4&={ \begin{pmatrix} 4 \\ 0 \end{pmatrix} } { x }^{ 4-0}{y}^{0} + { \begin{pmatrix} 4 \\ 1 \end{pmatrix} } { x }^{ 4-1}{ y }^{1} + { \begin{pmatrix} 4 \\ 2 \end{pmatrix} } { x }^{ 4-2 }{ y }^{ 2 } + { \begin{pmatrix} 4 \\ 3 \end{pmatrix} } { x }^{ 4-3 }{ y }^{ 3 } + { \begin{pmatrix} 4 \\ 4 \end{pmatrix} } { x }^{ 4-4 }{ y }^{ 4 } \\ &={ \begin{pmatrix} 4 \\ 0 \end{pmatrix} } { x }^{ 4} + { \begin{pmatrix} 4 \\ 1 \end{pmatrix} } { x }^{ 3}{ y } + { \begin{pmatrix} 4 \\ 2 \end{pmatrix} } { x }^{ 2 }{ y }^{ 2 } + { \begin{pmatrix} 4 \\ 3 \end{pmatrix} } { x }^{ 1 }{ y }^{ 3 } + { \begin{pmatrix} 4 \\ 4 \end{pmatrix} } { y }^{ 4 } \end{aligned}$

Recall that ${ \begin{pmatrix} 4 \\ 0 \end{pmatrix} }$ and ${ \begin{pmatrix} 4 \\ 4 \end{pmatrix} }$are both equivalent to 1, as there is only one way to choose either $0$ or $4$ objects from among $4$. Therefore, we have:

 $\displaystyle = { x }^{ 4} + { \begin{pmatrix} 4 \\ 1 \end{pmatrix} } { x }^{ 3}{ y } + { \begin{pmatrix} 4 \\ 2 \end{pmatrix} } { x }^{ 2 }{ y }^{ 2 } + { \begin{pmatrix} 4 \\ 3 \end{pmatrix} } { x }{ y }^{ 3 } + { y }^{ 4 } $

Now we must evaluate each of the remaining combinations: 

$\displaystyle \begin{pmatrix} 4 \\ 1 \end{pmatrix} = \frac { 4! }{ 1!(4-1)! } = \frac { 4! }{ 1!3! } = 4$

$\displaystyle \begin{pmatrix} 4 \\ 2 \end{pmatrix} = \frac { 4! }{ 2!(4-2)! } = \frac { 4! }{ 2!2! } = 6$ $$

$\displaystyle \begin{pmatrix} 4 \\ 3 \end{pmatrix} = \frac { 4! }{ 3!(4-3)! } = \frac { 4! }{ 3!1! } = 4$

Substituting these integers into the expansion, we have:

$\displaystyle (x+y)^4 = { x }^{ 4} + 4 { x }^{ 3}{ y } + 6 { x }^{ 2 }{ y }^{ 2 } + 4 { x }{ y }^{ 3 } + { y }^{ 4 } $