How to solve card related combination problem A deck of cards has 4 suits: diamonds, hearts, clubs, and spades. The suits of diamonds and hearts are both red and the suits of clubs and spades are both black. Each suit has the following denominations: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, and King. The Jacks, Queens and Kings are also called face cards.

Question: In how many ways 2 cards can be drawn such that one card is from red face cards and the other is a black card.    

I know this is a combination problem, and I have tried solving it by taking (26!/1!(26-1)!) * (26!/1!(26-1)!). My main concern is whether or not I have to account for whether 2 black cards can be drawn with 0 red cards or 0 black cards drawn with 2 red cards. Thank you for any advice.
No you don't need to take cases 0 red, 2 black and 2 red, 0 black. As from question it is cleared that we need 1 black and 1 red face card.
If the order doesn't matter (drawing the red face after or before the black card is the same) then you have 

$$\dfrac{(\text{total of red faces})\times (\text{total black cards})}{2} $$

where the 2 comes from the fact that the order doesn't matter. There are 52 cards, 26 of each color and 3 faces every suit, therefore 6 faces of each color. Hence, the numbers are:

$$\dfrac{6 \times 26}{2} = 78 $$