I was just going back and refreshing my knowledge on binomial coefficients/probability and I was trying to calculate some poker odds since I would be able to look them up and verify my attempts.
I have run into an issue when trying to calculate how many off-suit hands there are for an opening deal in Texas Hold'em.
The way I'm attempting it is that I pick 2 suits of the 4 available $(\binom{4}{2} = 6)$ and then pick 2 ranks of the 13 ($\binom{13}{2} = 78$) to ensure that I don't get any pairs or any suited cards.
I multiply the results together $\binom{4}{2} \times \binom{13}{2} = 6 \times 78 = 468$
Only problem is I'm missing half since the number of possible unsuited hands you can get is apparently 936 according to https://en.wikipedia.org/wiki/Texas_hold_%27em_starting_hands
Where am I going wrong with my logic here?
When you pick two suits and two ranks you have missed matching the suits with the ranks. If you pick Spades and Hearts, then $2$ and $3$, you could have $2S, 3H$ or $2H, 3S$. That is another factor of $2$.