Binomial Expansion refers to expanding an expression that involves two terms added together and raised to a power, i.e. $(x+y)^n$. Before learning how to perform a Binomial Expansion, one must understand factorial notation and be familiar with Pascal’s triangle.

## Factorial Notation

When you see an exclamation mark following a number in mathematics it is known as a factorial. For example, 6! is said ‘6 factorial’ and you multiply all of the positive integers less than 6 together: $6!=6\times 5\times 4 \times 3 \times 2 \times 1=720$

Here are some more examples: $8!=8\times 7\times 6\times 5\times 4 \times 3 \times 2 \times 1=40,320$ $4!\times 3!=4\times 3\times 2 \times 1\times 3\times 2 \times 1=24\times 6=144$ $9!\div 5!=\frac{9\times 8\times 7\times 6\times 5\times 4 \times 3 \times 2 \times 1}{5\times 4 \times 3 \times 2 \times 1}=9\times 8\times 7\times 6=3,024$

## Pascal’s triangle

Pascal’s triangle is the pyramid of numbers where each row is formed by adding together the two numbers that are directly above it:

0th row:                                         1
1st row:                                       1  1
2nd row:                                   1  2  1
3rd row:                                1   3   3   1
4th row:                             1   4    6    4   1
5th row:                          1   5   10   10   5   1
6th row:                      1    6   15   20   15   6   1
7th row:                   1   7   21   35   35   21   7   1

The triangle continues on this way, is named after a French mathematician named Blaise Pascal (find out more about Blaise Pascal) and is helpful when performing Binomial Expansions.

Notice that the 5th row, for example, has 6 entries. Like the 0th row, the first entry in any one row is the 0th entry.  Consider the first 15 in the 6th row, we call this $^{6}C_{2}$, pronounced ‘6 choose 2’. This can also be written as ${6}\choose{2}$. In general, we write $^{n}C_r$ or ${n}\choose {r}$ and is calculated as $^{n}C_r=\left(\begin{array}{c}n\\r\end{array}\right)=\frac{n!}{(n-r)!r!}$

This comes from summing all the terms above a given entry and simplifies to a fraction with factorials. $^{n}C_r$ can be thought of as the number of combinations of putting r balls in n buckets. It is also the number of times you get an $x^ry^{n-r}$ term in the expansion of $(x+y)^n$. Hence, this is why Pascal’s triangle is useful in Binomial Expansion. Note that there is a button on your calculator for working out ${n}\choose{r}$ – you don’t necessarily need to calculate the individual factorials. You might also notice that ${{n}\choose{0}}={{n}\choose{n}}=1$ and ${{n}\choose{1}}={{n}\choose{n-1}}=n$ always.

## Binomial Expansion

Suppose now that we wish to expand $(x+y)^n$, i.e. find the Binomial Expansion. In the simple case where n is a relatively small integer value, the expression can be expanded one bracket at a time. See Examples 1 and 2 below.

### Example 1

Expand $(x+y)^3.$

### Example 2

Using Example 1 expand $(x+y)^4.$

Expanding $(x+y)^n$ by hand for larger n becomes a tedious task. The Edexcel Formula Booklet provides the following formula for binomial expansion: $(a+b)^n=a^n+\left(\begin{array}{c}n\\1\end{array}\right)a^{n-1}b+\left(\begin{array}{c}n\\2\end{array}\right)a^{n-2}b^2+...+\left(\begin{array}{c}n\\r\end{array}\right)a^{n-r}b^r+...+b^n$

where $\left(\begin{array}{c}n\\r\end{array}\right)=\frac{n!}{(n-r)!r!}$

(see above) for when $n\in{\mathbb N}$, i.e for when n is a positive integer.  Directly substituting x in place of a and y in place of b results in finding the expansions for larger n. Usually only the first few terms are required – see Example 3. You may substitute other expressions or numbers for a and b and you may be asked for ascending or descending powers of a particular variable. See Example 4 – you will notice that when there are also coefficients inside the brackets, the coefficients in the expansion change dramatically from those given in Pascal’s triangle.

### Example 3

Find the first three terms in the expansion $(x+y)^8$.

### Example 4

Find the first three terms, in descending powers of x, of the binomial expansion of $(2x+4)^5$.

Now check out the further examples below to see what an exam question might look like.

## Relationship to Probabilities

Consider a binomially distributed random variable with n trials and probability of success p – see Binomial Distribution. If we require r of the trials to be successful (probability $p^r$) we require the remaining n-r trials to be unsuccessful (probability $(1-p)^{n-r}$). The number of combinations in which there can be r successes out of n trials is ${n}\choose{r}$ (see above). Finally, the associated probability is given by $P(X=r)={{n}\choose{r}}p^r(1-p)^{n-r}$

when $X\sim B(n,p)$  as seen on the Binomial Distribution page.

## More Binomial Expansion Examples

### Sample Exam Question

1. Given that ${{8}\choose{2}}=\frac{8!}{2!q!}$, write down the value of q.
2. Given that the coefficient of $x^2$ in the expansion of $\left(p-\frac{x}{8}\right)^8$ is 28. find the value of p
3. Using the first three terms of a binomial expansion, estimate the value of $1.995^8$.

1. The formula for ‘n choose r’ is given by ${{n}\choose{r}}=\frac{n!}{r!(n-r)!}$. Setting n=8 and r=2 gives the missing term n-r=6 and so q=6.
2. One can perform the full expansion up to the $x^2$ term or notice that only the coefficient of $x^2$ is required. That is, the coefficient when the term ${{8}\choose{2}}p^6\left(-\frac{x}{8}\right)^2$ is simplified. Notice that a and b are interchangeable in the formula and are chosen so that the power of x is 2. The coefficient is thus $28\times p^6\times \frac{1}{64}$ and equal to 28 giving p=2.
3. Find the first 3 terms in the binomial expansion using p=2: $\left(2-\frac{x}{8}\right)^8\approx 2^8-{{8}\choose{1}}\times 2^7\times-\frac{x}{4}+28 x^2=256-256x+28x^2$ Note that we already knew the coefficient of the $x^2$ term. Using this expansion suggests that we should choose x so that $2-\frac{x}{8}=1.995$, that is, $0.005=\frac{x}{8}$ or x=0.04. Substituting x=0.04 into the expansion gives $(1.995)^8\approx256-256\times 0.04+28\times 0.04^2=245.8048$  The actual answer to 7 decimal places, using a calculator, is 250.9245767, so not a great approximation.

## Binomial Expansion Board Question Note that C2 was a module when A-Level Maths was delivered in a modular fashion. Binomial Expansion is now an AS-Level topic.

## Statistics Example

Consider the binomially distributed random variable $X\sim B(7,x)$. Find the probability $P(X=3)$ in terms of x. Write your answer as a polynomial in x.

Using the formula: $P(X=3)={{7}\choose{3}}x^3(1-x)^4$

We can use Example 2 above to expand $(1-x)^4$: $P(X=3)=35x^3\left(1-4x+6x^2-4x^3+x^4\right)$ $=35x^3-140x^4+210x^5-140x^4+35x^3$

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