n choose 2 is
n(n - 1) / 2
This is
n2 / 2 - n/2
We can see that n(n-1)/2 = Θ(n2) by taking the limit of their ratios as n goes to infinity:
limn → ∞ (n2 / 2 - n / 2) / n2 = 1/2
Since this comes out to a finite, nonzero quantity, we have n(n-1)/2 = Θ(n2).
More generally: n choose k for any fixed constant k is Θ(nk), because it's equal to
n! / (k!(n - k)!) = n(n-1)(n-2)...(n-k+1) / k!
Which is a kth-degree polynomial in n with a nonzero leading coefficient.
Hope this helps!