Understanding calculation of collisions in hash-functions
I suspect you are misrepresenting what your professor actually said.

Since I'm not certain exactly what he said, here is how I would explain it:


With 1024 outputs, there are $\binom{1024}{2} \approx 1024^2/2$ pairs of outputs.
For each pair of output, that pair has a $2^{-12}$ probability of being a collision (that is, those two outputs being exactly the same).
Hence, the expected number of collisions would be about $1024^2/2 \times 2^{-12} = 128$


The exact expected number would depend how you count a multiway collision (where 3 or more outputs have the same value); it turns out that, if you count it right, $\binom{1024}{2} 2^{-12}$ is the correct answer.