Using the hypergeometric distribution shows how to put #2 in nCr form similar to #1.

After the draw of the first marble, the next two draws are from a hypergeometric distribution with 1 object of the first type and 3 objects of the second type. (The hypergeometric is used with sampling without replacement, which this is.) Using the formula for selecting 2 objects of the second type from that distribution, the probability is

\(\displaystyle \Large \frac{{1 \choose 0}{3 \choose 2}}{{4 \choose 2}}\).

And the total probability including the first draw is

\(\displaystyle \Large \frac{2}{5} \times \frac{{1 \choose 0}{3 \choose 2}}{{4 \choose 2}}\).

Evaluating that yields the same probability as pka.

If you were not taught the hypergeometric distribution, it's possible to derive that formula using a direct argument (but no one actually would).