Large round brackets in equation I cannot recall what the large () brackets mean - Google seems to be full of links on how to create them but not what they actually are or how to resolve them.
$
\left(
\begin{array}{l}
n\\
2
\end{array}
\right)r^2
$
Can someone please clarify? In this instance n = 6 and r = 0.05 if that helps.
 A: The binomial coefficient 
$$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$ 
is the number of ways of selecting $k$ elements from $n$ elements when order does not matter (the number of subsets with $k$ elements in an $n$ element set).
The number $n!$, read "$n$ factorial," is defined recursively as follows:


*

*$1! = 1$

*$n! = n(n - 1)!$ for $n \geq 1$


If you substitute $1$ in the definition for $n!$, you obtain 
\begin{align*}
1! & = 1(1 - 1)!\\
1 & = 1 \cdot 0!\\
1 & = 0!
\end{align*}
For positive integers, $n!$ is the product of the first $n$ positive integers.  For instance,
\begin{align*}
6! & = 6 \cdot 5!\\
   & = 6 \cdot 5 \cdot 4!\\
   & = 6 \cdot 5 \cdot 4 \cdot 3!\\
   & = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2!\\
   & = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1!\\
   & = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1
\end{align*}
Hence, 
$$\binom{6}{2} \cdot 0.05^2 = \frac{6!}{2!4!} \cdot 0.0025 = \frac{6 \cdot 5 \cdot 4!}{2 \cdot 1 \cdot 4!} \cdot 0.0025 = 15 \cdot \frac{1}{400} = \frac{3}{80}$$
A: it means the number of combinations possible when you pick 2 items out of n, that is n(n-1)/(2*1)
