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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.

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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.

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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 $6\cdot26=156$ combinations.

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

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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.

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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:

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

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  • $\begingroup$ Why the $2$ in the denominator? $\endgroup$ – prog_SAHIL Aug 5 '18 at 16:08

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