# What Is n Choose 2?

Solving for $n$ choose $2$ means finding the number of ways of choosing $2$ items from a group with a population of $n$. This is a problem that uses combination formula. However, after the derived formula for $n$ choose $2$ after using the combination formula, we observe that it is an expression for something else. Read this guide to know what is $n$ choose $2$ equivalent to.

## What Is $n$ Choose 2?

The expression $n$ choose $2$, in symbol $\binom{n}{2}$, is the sum of the first consecutive $n-1$ integers. That is, the sum of $1,2,3,\dots,n-1$ is equal to $n$ choose $2$. In mathematical notation, we express it as:

\begin{align*}
1+2+\dots+n-1= \sum_{i=1}^{n-1} i=\binom{n}{2}.
\end{align*}

Using the formula for summation, we know that the sum of the first $n$ integers is $\dfrac{n(n+1)}{2}$. Thus, we have
\begin{align*}
\sum_{i=1}^{n-1} i=\dfrac{(n-1)(n-1+1)}{2}=\dfrac{(n-1)n}{2}=\binom{n}{2}.
\end{align*}

Hence, $n$ choose $2$ is equal to $\dfrac{n(n-1)}{2}$.

## Combination Counting Technique

Combination is one of the counting techniques that is used when we want to know how many possible ways can we pick $r$ objects from a group with a total of $n$ objects, without giving importance to the order.

### Example

For instance, we want to know the number of ways of selecting three letters from the letters $A,B,C,D,E$. Using a manual enumeration and grouping of letters, we get the following groupings of letters:
\begin{align*}
\end{align*}

Note that we no longer put $CEA$ because it is the same as $ACE$ since the order does not matter. From this, we can see that we are able to list down 10 groups of letters. Thus, there are 10 possible ways of forming a group of three letters from a group of five letters.

## What Is the Combination Formula?

The combination formula is a formula that computes the number of unordered groups of $r$ objects from $n$ objects. This can also be interpreted as the number of combinations of $n$ objects taken $r$ at a time, denoted by $\binom{n}{r}$. The formula for combination is given by
\begin{align*}
\binom{n}{r}=\dfrac{n!}{\left(n-r\right)!r!}.
\end{align*}

The notation $\binom{n}{r}$ can also be read as $n$ choose $r$. The combination formula is used to ease up solving problems involving combination counting techniques and probabilities so that we don’t have to enumerate all the possible combinations. The formula is a very helpful tool, especially for large values of $n$ and $r$.

In this article, we evaluate $n$ choose 2, denoted as $\binom{n}{2}$. That is, we need the total number of groups of two elements that could be formed from $n$ objects.

### Recall: Factorial Notation

Note that the notation $!$ denotes factorial. So, the expression $n!$ is read as $n$ factorial and is solved using the formula \begin{align*} n!=n\times\left(n-1\right)\times\left(n-2\right)\times\dots\times2\times1. \end{align*} For example, $5!$ is $120$ because \begin{align*} 5!=5\times4\times3\times2\times1=120. \end{align*}

### Example: Evaluate 4 choose 3

We rewrite 4 choose 3 into its notation, $\binom{4}{3}$. We use the combination formula to evaluate $\binom{4}{3}$, where $n=4$ and $r=3$. Then, we have: \begin{align*} \binom{4}{3}&=\dfrac{4!}{\left(4-3\right)!3!}\\ &=\dfrac{4!}{1!3!}\\ &=\dfrac{\left(4\times3\times2\times1\right)}{\left(1\times\left(3\times2\times1\right)\right)}\\ &=\dfrac{4}{1}\\ &=4. \end{align*} Hence, 4 choose 3 is equal to 4. This implies that there are exactly only four possible ways of picking 3 elements from a group of 4 objects.

## Evaluating $n$ Choose 2

Evaluating $n$ choose 2 will give us the formula
\begin{align*}
\binom{n}{2}=\dfrac{n\left(n-1\right)}{2}.
\end{align*}

We use the combination formula to derive the $n$ choose 2 formula. Plugging in $r=2$ in the combination formula, we have
\begin{align*}
\binom{n}{2}&=\dfrac{n!}{\left(n-2\right)!2!}.
\end{align*}

Note that $n!$ can be expressed as
\begin{align*}
n!=n\times\left(n-1\right)\times\left(n-2\right)!.
\end{align*}

Thus, we have
\begin{align*}
\binom{n}{2}&=\dfrac{n!}{\left(n-2\right)!2!}\\
&=\dfrac{\left(n\times\left(n-1\right)\times\left(n-2\right)!\right)}{\left(n-2\right)!2!}\\
&=\dfrac{n\left(n-1\right)}{2!}\\
&=\dfrac{n\left(n-1\right)}{2}.
\end{align*}

Note that, since $n$ is a variable, then we cannot directly solve or express $\binom{n}{2}$ as a number. Hence, we can only form the corresponding formula in evaluating n choose 2.

We can now use this $n$ choose 2 simplified formula to solve for problems involving choosing 2 objects from a number of objects without using the initial combination formula.

### Example

• What is 6 choose 2?

Since $n$ choose 2 is the sum of the first $n-1$ integers, then 6 choose 2 is the sum of the first 5 integers. That is,
\begin{align*}
\binom{6}{2} = 1+2+3+4+5.
\end{align*}

Letting $n=6$, and using the formula, we have
\begin{align*}
\binom{6}{2} = \dfrac{6(6-1)}{2}=\dfrac{(6)(5)}{2}=15.
\end{align*}

We verify this by taking the sum of 1, 2, 3, 4, 5. Thus, we have
\begin{align*}
1 + 2 + 3 + 4 + 5= 15.
\end{align*}

Hence,
\begin{align*}
\binom{6}{2} = 1+2+3+4+5 = 15.
\end{align*}

### Example: Solve for 5 Choose 2

To evaluate 5 choose 2, we let $n=5$, then proceed to use the formula we obtained in the previous section. Thus, we have \begin{align*} \binom{5}{2}&=\dfrac{5\left(5-1\right)}{2}\\ &=\dfrac{5(4)}{2}\\ &=\dfrac{20}{2}\\ &=10. \end{align*} Therefore, $\binom{5}{2}=10$.

### Example: Evaluate 12 Choose 2

We take $n=12$ in order to evaluate $\binom{12}{2}$. Then, we apply it to the formula for $n$ choose 2. So, we have: \begin{align*} \binom{12}{2}&=\dfrac{12\left(12-1\right)}{2}\\ &=\dfrac{12(11)}{2}\\ &=\dfrac{12}{2} \left(11\right)\\ &=6\left(11\right)\\ &=66. \end{align*} Thus, $12$ choose $2$ evaluated is equal to $66$.

## Importance of n Choose 2

The importance of solving $n$ choose 2 relates to the importance of arithmetic series. From, the formula we obtained for $n$ choose 2, we can further express it as: \begin{align*} \binom{n}{2}&=\dfrac{n\left(n-1\right)}{2}\\ &=\dfrac{\left(n-1\right)n}{2}\\ &=\dfrac{\left(n-1\right)\left(\left(n-1\right)+1\right)}{2}. \end{align*} Recall that the sum of the first $k$ integers is given by \begin{align*} 1+2+3+\dots+k=\dfrac{k\left(k+1\right)}{2}. \end{align*} Notice that the formula for $n$ choose 2 is the sum of the first consecutive $n-1$ integers. Moreover, the binomial coefficient $n$ choose 2 indicates the $(n-1)th$ triangular number.

## Sum of n Choose 2

Another property of $n$ choose 2 is the sum of these coefficients can be generalized by a single binomial coefficient. The sum of $n$ choose 2 is given by \begin{align*} \sum_{i=2}^{n}\binom{i}{2}&=\binom{2}{2}+\binom{3}{2}+\binom{4}{2}+\dots+\binom{n}{2}\\ &=\binom{n+1}{3}. \end{align*}

### Example

Find the sum of the first ten terms of the sequence $\binom{n}{2}$. To solve this, instead of individually solving for $\binom{2}{2}$,$\binom{3}{2}$, and so on. We can just use the simplified formula for the sum of $n$ choose 2. Note that since we are solving for the sum of the first 10 terms, and the first term is $\binom{2}{2}$, then $n=11$. Thus, we have: \begin{align*} \sum_{i=2}^{n=11} \binom{i}{2}&=\binom{11+1}{3}\\ &=\binom{12}{3}\\ &=\dfrac{12!}{\left(12-3\right)!3!}\\ &=\dfrac{\left(12\times11\times10\times9!\right)}{\left(9!\right)3!}\\ &=\dfrac{\left(12\times11\times10\right)}{3!}\\ &=\dfrac{12}{6} \left(11\times10\right)\\ &=2\times11\times10\\ &=220. \end{align*} Therefore, the sum of the first ten terms of the sequence $\binom{n}{2}$ is $220$.

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