1. The problem statement, all variables and given/known data In a 52 card deck, if you draw 5 cards find the probability of drawing exactly one ace. 2. Relevant equations (n k) = n!/k!(n-k)! P(A) = |A|/|S| 3. The attempt at a solution So I took the logic of a different example we had that stated it like this - If we have 5 options but only 1 of them can be an ace then we have (5 1). This gives a very very unrealistic problem as the sample space is (52 5) yielding me 5/2598960 to equal roughly 1.62*10^-6 So I took a different logic stating that if we have 4 cards that are aces then 48 of them are not. Then we may choose 4 of these cards for (48 4). My question is which one of these or are neither of them a very profficient way to approach these? I've seen it put where you can logically construct these situations using something to the effect of: (n1 k1)(n2 k2) and this would produce the appropriate number for your |A| because you construct it through multiplecation. ie (48 1) = (12 1)(4 1).