## Identifying Binomial Coefficients

In Counting Principles, we studied combinations. In the shortcut to finding ${\left(x+y\right)}^{n}$, 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 $\left(\begin{array}{c}n\\ r\end{array}\right)$ instead of $C\left(n,r\right)$, but it can be calculated in the same way. So

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

The combination $\left(\begin{array}{c}n\\ r\end{array}\right)$ is called a binomial coefficient. An example of a binomial coefficient is $\left(\begin{array}{c}5\\ 2\end{array}\right)=C\left(5,2\right)=10$.

### A General Note: Binomial Coefficients

If $n$ and $r$ are integers greater than or equal to 0 with $n\ge r$, then the binomial coefficient is

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

### Is a binomial coefficient always a whole number?

Yes. Just as the number of combinations must always be a whole number, a binomial coefficient will always be a whole number.

### Example 1: Finding Binomial Coefficients

Find each binomial coefficient.

1. $\left(\begin{array}{c}5\\ 3\end{array}\right)$
2. $\left(\begin{array}{c}9\\ 2\end{array}\right)$
3. $\left(\begin{array}{c}9\\ 7\end{array}\right)$

### Solution

Use the formula to calculate each binomial coefficient. You can also use the ${n}_{}{C}_{r}$ function on your calculator.

$\left(\begin{array}{c}n\\ r\end{array}\right)=C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}$
1. $\left(\begin{array}{c}5\\ 3\end{array}\right)=\frac{5!}{3!\left(5 - 3\right)!}=\frac{5\cdot 4\cdot 3!}{3!2!}=10$
2. $\left(\begin{array}{c}9\\ 2\end{array}\right)=\frac{9!}{2!\left(9 - 2\right)!}=\frac{9\cdot 8\cdot 7!}{2!7!}=36$
3. $\left(\begin{array}{c}9\\ 7\end{array}\right)=\frac{9!}{7!\left(9 - 7\right)!}=\frac{9\cdot 8\cdot 7!}{7!2!}=36$

### Analysis of the Solution

Notice that we obtained the same result for parts (b) and (c). If you look closely at the solution for these two parts, you will see that you end up with the same two factorials in the denominator, but the order is reversed, just as with combinations.

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

### Try It 1

Find each binomial coefficient.

a. $\left(\begin{array}{c}7\\ 3\end{array}\right)$
b. $\left(\begin{array}{c}11\\ 4\end{array}\right)$

Solution