Why is ${n \choose 2}$ a $\Theta(n^2)$ computation?
$\Theta$-notation more generally refers to the growth of a function. In terms of computational complexity this function can be done in $O(1)$ as:

$$\frac{n!}{(n-2)!2!} = \frac{n(n-1)}{2!} = \frac{n(n-1)}{2}$$

Since there are a finite number of multiplications and multiplication is sometimes treated as $O(1)$ this has constant computational time. But, the author is not referring to the computational complexity of this function. Instead he is referring to the growth of the function:

$$\frac{n(n-1)}{2} = \frac{n^2}{2} - \frac{n}{2} \in \Theta(n^2)$$