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.
- club that is not an ace, ace of clubs
- ace of clubs, ace that is not a club
- 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?
The (fanmade?) book solutions says:
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?