Vtyjkyuk

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.

There are $6$ red face cards, and $26$ black cards. The categories don't overlap, and we need one from each ("one... and the other..."). Thus there are $6cdot26=156$ combinations.

If order matters, then this number is doubled to $312$ valid drawings.

Picking one red face card i.e. $C^6_1=6$

Picking a black card i.e. $C^26_1=26$

Total ways to pick $ = 6 cdot 26 = 156$

You don't need to worry about any overlaps as both the sets i.e. Red face cards and Black cards have no intersection.

If the order doesn't matter (drawing the red face after or before the black card is the same) then you have

$$dfrac(texttotal of red faces)times (texttotal black cards)2 $$

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:

$$dfrac6 times 262 = 78 $$