Attached is a photo of the problem I'm having a hard time with.
We are trying to calculate the likelihood of drawing an ace given 10 or more draws with no aces drawn.
We are given two methods of evaluation, I'm trying to make sense of the counting method, in particular:
I'm trying to justify the 41!
It is clear to me why we choose the falling factorial 48×...×40 for the first 9 cards, but I've been trying to reason through the 41!, and I don't feel solid in my justifications.
I was thinking we have 10 cards chosen, so (52-10)! Permutations for the remaining cards but that gives us 42! And we only chose 9 cards, not 10.
What am I missing in terms of counting the event space?
You found a mistake in the book and Your reasoning is correct . Here however "10 or more cards before the first ace appears" is obviously interpreted as "the first $9$ cards are no aces", quite incorrectly if You take "before" literally. In this case after drawing the first $9$ cards, that are no aces, there are $43$ cards left and ideed $$P(A)=\frac{48\cdot74\cdot.46\cdot\cdot\cdot\cdot40\cdot 43!}{52!}\approx 0.4559$$ as You can easily check.( Actually even the equation $P(A)=48\cdots40\cdot41!/52!=0.4559$ is wrong computationally, this expression gives a much smaller value.)