# first card club and second card ace?

Larsen and Marx

Suppose that two cards are drawn—in order— from a standard 52-card poker deck. In how many ways can the first card be a club and the second card be an ace?

I think there are 3 cases.

1. club that is not an ace, ace of clubs
2. ace of clubs, ace that is not a club
3. club that is not an ace, ace that is not a club

Case 1: $12 \times 1=12$

Case 2: $1 \times 3=3$

Case 3: $12 \times 3=36$

Thus we have $12+3+36 = 51$.

Am I wrong?

Case 1: Ace of clubs is one of those cards

Case 2: Ace of clubs is not one of those cards

Case 1: $2 \times 1 \times 36=72$

Case 2: $2 \times 3 \times 12=72$

Thus, we have $72+72=144$

How do you come up with those numbers? I'm guessing for case 1 the 36 comes from $12 \times 3$ or $52-4-13+1$ (all cards - 4 aces - 13 clubs + ace of clubs) and the 2 comes from 2 places to draw the ace of clubs?

• It seems to me that the book solution is answering a different question. Something like: Two cards are drawn. Exactly one of them is an ace, and exactly one of them is a club. In how many ways can we draw the cards so that these conditions are met. – Jyrki Lahtonen Jun 13 '16 at 10:42

Your solution is correct. However, it could be reduced to two cases.

1. The first card is a club that is not the ace of clubs and the second card is an ace.
2. The first card is the ace of clubs and the second card is an ace that is not the ace of clubs.

Case 1: There are twelve ways of selecting a club that is not the ace of clubs in the first draw and four ways to select one of the four available aces in the second draw. Hence, there are $12 \cdot 4 = 48$ possible selections in the this case.

Case 2: There is one way to select the ace of clubs in the first draw and three ways to select one of the three remaining aces in the second draw. Hence, there are $1 \cdot 3 = 3$ possible selections in the second case.

Total: $12 \cdot 4 + 1 \cdot 3 = 48 + 3 = 51$

The authors of your text seem to be answering a different question, namely, in how many ways can exactly one ace and exactly one club be drawn if two cards are selected from a deck?

They do not take into account the fact that the first card is a club and the second card is an ace. They are allowing the cards to be drawn in either order.

Ace of clubs is one of those cards:

Select the ace of clubs, select one of the $36$ cards that is neither an ace nor a club, then multiply by $2$ to account for the two possible orders in which the cards could be selected.

Ace of clubs is not one of those cards:

Select one of the other three aces, select one of the twelve clubs that is not the ace of clubs, then multiply by $2$ to account for the two possible orders in which the cards could be selected.

I think both solutions are wrong.

There are $13$ clubs and $4$ aces, for a total of $13\cdot 4=52$ combinations -- except that the one where both cards are the ace of clubs is impossible, and we need to subtract that. So the true answer ought to be $13\cdot 4-1=\underline{\underline{51}}$.

(Actually this is the same result as you would get with your method if you avoid making making arithmetic errors in the final addition).

• Thanks Henning Makholm ^-^ – BCLC Jun 14 '16 at 20:44