In a deck of cards that are turned up one at a time until the first A appears. Is the next card more likely to be the A of spades or the 2 of clubs? This is  Example 5j, from Sheldon Ross's First Course in Probability 8th ed, page 38. I don't understand why the following is true.
Solution.To determine the probability that the card following the first ace is the aceof spades, we need to calculate how many of the (52)! possible orderings of the cardshave the ace of spades immediately following the first ace. To begin, note that each ordering of the 52 cards can be obtained by first ordering the 51 cards different from the ace of spades and then inserting the ace of spades into that ordering.
I don't see how this (sentence in italics) can be true. For example if we have $S=\{ 1, 2, 3\} $ the number of orderings that can be obtained are $3!=6$. Following the solution's reasoning we could calculate the orderings for $S$ by ordering the cards different form $3$ and then inserting in into that ordering, that is $ 2!$ .
What am I missing? perhaps the sentence in italics does not mean what I think it does?
Also, the solution given is a probability of $ \frac{1}{52} $ for both, I understand why but I have a different solution that also seems valid:
My solution

*

*Ordering in which the card following the first ace is the ace of spades;
We have 3 other aces so we put $A_i A_s  $,  with $i = c, d, h $,  together as one unit  and count the number of permutations  $ = 51! $. As we have three of these such pairs $$ P(N_a) = \dfrac{3\cdot 51!}{52!} $$


*Ordering in which the card following the first ace is the two of clubs By a similar argument we put $A_i A_s  $,  with $i = c, d, h, s$, so
$$ P(N_c) = \dfrac{4\cdot 51!}{52!} $$
Can someone tell me what is the error in this reasoning?

Given that the ﬁrst ace is the 20th card to appear:
This suggests no Ace was chosen in the first 19 card flips.  No other information was given, so every other card is subject to random choice in the first 19 card flips.  On the 20th flip we are told an Ace is flipped.  This gives the equation for P(A)
$P(A) = \binom{48}{19} * \binom{4}{1}$
On the 21st flip we are told an Ace of Spades is flipped. This means there are only 3 Aces to be randomly selected on the 20th flip.  On the 21st flip there are 32 cards to chose from and we need the probability that the Ace of Spades is chosen.
$P(BA) = \binom{48}{19} * \binom{3}{1} * \dfrac1{32}$
So the final equation is
$P(B \mid A) = \dfrac{\binom{48}{19} * \binom{3}{1} * \dfrac1{32}}{\binom{48}{19} * \binom{4}{1}}$
On the 21st flip we are told the Two of Clubs is flipped. This means in addition to the 4 Aces, the Two of Clubs is not chosen in the first 19 flips.  This leaves 47 cards to be chosen at random in the first 19 flips.  On the 20th flip one of four Aces are chosen.  On the 21st flip the Two of Clubs is chosen.
$P(CA) = \binom{47}{19} * \binom{4}{1} * \dfrac1{32}$
So
$P(C \mid A) = \dfrac{\binom{47}{19} * \binom{4}{1} * \dfrac1{32}}{\binom{48}{19} * \binom{4}{1}}$
