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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*}
ABC,ABD,ACD,ACE,ADE,BCD,BCE,BDE,CDE.
\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 $220$.

Simplifying n Choose 3

Similar to $n$ choose 2, we can also derive a simpler formula for $n$ choose 3 so that we can also have a simplified expression for the sum of $n$ choose 2. Using the combination formula for $n$ choose 3, we have: \begin{align*} \binom{n}{3}&=\dfrac{n!}{\left(n-3\right)!3!}\\ &=\dfrac{\left(n\times\left(n-1\right)\times\left(n-2\right)\times\left(n-3\right)!\right)}{\left(n-3\right)!3!}\\ &=\dfrac{n\left(n-1\right)\left(n-2\right)}{3!}\\ &=\dfrac{n\left(n-1\right)\left(n-2\right)}{6}. \end{align*} Thus, $n$ choose 3 can be simply expressed as $\binom{n}{3}=\dfrac{n\left(n-1\right)\left(n-2\right)}{6}.

Example: What is 7 choose 3?

We first solve 7 choose 3. Using the formula we derived earlier, we let $n=7$. Then, we have: \begin{align*} \binom{7}{3}&=\dfrac{7\left(7-1\right)\left(7-2\right)}{6}\\ &=\dfrac{7\left(6\right)\left(5\right)}{6}\\ &=7(5)\\ &=35. \end{align*} Thus, 7 choose 3 is 35. We can also $\binom{7}{3}$ as: \begin{align*} \binom{7}{3}=\binom{6+1}{3}. \end{align*} Hence, 7 choose 3 is also the sum of the first 5 terms of the sequence n choose 2.

Conclusion

In this article, we focused on evaluating $n$ choose 2, its equivalence and importance, and some of the consequences of its properties. We list down a summary of the vital points in this discussion.

  • $n$ choose 2 is the sum of the first consecutive $n-1$ integers.
  • The simplified formula for $n$ choose 2 is given by $\binom{n}{2}=\dfrac{n\left(n-1\right)}{2}$.
  • The sum of the first $n-1$ integers is equal to $n$ choose 2.
  • The sum of the sequence generated by $n$ choose 2 is $\binom{n+1}{3}$.
  • The simplified formula for $n$ choose 3 is given by $\binom{n}{3}=\dfrac{n\left(n-1\right)\left(n-2\right)}{6}$.

The combination counting techniques are used in determining binomial coefficients and could be further explored to learn more simplified patterns or formulas for the coefficients. The connection between summation and binomial coefficients can also be looked into as established by the expression $n$ choose 2.

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