How many ways to pick two cards without replacement?

Question

From the deck of 52 cards, in how many ways two cards can be picked such that the first card is spades and second card is not a face card.

My try: Here we can have two cases.

Case 1. First card is not a face card of spade and second card is any card but not face card.

Case 2. First card is a face card and second card is any card but not face card.

Now, I am not getting, how to count the cards. Any help would be appreciated.

Answer 1

Let us consider all the relevant cards in the deck:

• There are 13 spades (ace through to king)
• There are 40 non-face cards (i.e. every card other than the jack, queen and king of each suit)

There is, of course, overlap between these subsets. Specifically, there are 10 cards in the deck which are spades and are not face cards.

Let us consider the possibilities for the first card:

There are two cases. In case one, the card is one of 10 spades that is not a face card, thus it is in the overlap between the two subsets. In case two, the card is one of 3 face cards of spades, i.e. it is one of the Jack of Spades, Queen of Spades, or King of Spades.

In the first case, card one can be one of only 10 cards, yet card two can only be thirty nine of the forty cards from the second subset since card one is also in this subset (it is in the overlap). Therefore, the number of possible ways to pick two valid cards in case two is $10 \times 39 = 390$.

In the second case, the first card can be one of only three cards, while card two can be any of the forty cards from the second group. Thus for case two, in total, there are $3 \times 40 = 120$ possible ways to pick two valid cards.

Hence, overall, there are $120 + 390 = 510$ different ways to pick two cards out of a deck with the given constraints.