Ace Of Spades Right After An Ace and Two of Clubs Right After An Ace

Question

I came across this question

A Deck of 52 playing cards is shuffled and the cards are turned up and kept on the table. What is the probability that ace of spades comes right after an ace and 2 of clubs comes right after an ace?

I tried the following approach:

1. Sample space consists of $52!$ orderings of the cards.
2. Let an event A be all the orderings such that ace of spades is right next to an ace and two of clubs is right next to an ace.

lets denote the set of events described below as $B$. We visualize the 3 aces and denote their positions as $i, j, k (1 \leq i, j, k \leq 52)$
We can choose 3 slots in ${51\choose 3}$ ways (51 because we should not choose the $52^{nd}$ position as we need slots that have an empty slot right next to it. We can permute the 3 aces in these 3 slots in $3!$ ways. Out of these 3 slots, we can choose any 2 slots(for ace of spades and two of clubs) in ${3\choose 2} \cdot 2!(=P(3, 2))$. And the remaining 47 cards can be arranged in $47!$ ways.

Therefore, $$|B| = {51\choose 3} \cdot P(3, 2) \cdot 3! \cdot 47!$$

lets denote the set of events described below as $C$. We visualize the 3 aces with one ace in the $52^{nd}$ position. We can choose the remaining 2 slots in ${51\choose 2}$ ways. We can permute the 3 aces in these 3 slots in $3!$ ways. Out of these 2 slots, we can choose any 2 slots(for ace of spades and two of clubs) in $P(2, 2)$ ways. And the remaining 47 cards can be arranged in $47!$ ways. Therefore. $$|C| = {51\choose 2} \cdot P(2, 2) \cdot 3! \cdot 47!$$

Lets denote the set of events described below as $D$. We can choose from 50 positions for the 3 aces( we cannot choose the 51 and 52 positions because we need 2 empty slots right next to the ace), that is ${50 \choose 3}$ ways. The pair (ace of spades, 2 of clubs) can sit right next to either of these 3 slots in 3 ways. The remaining 47 cards can be arranged in $47!$ ways. Therefore, $$|D| = {50 \choose 3} \cdot 3! \cdot 3 \cdot 47!$$

$B$ and $C$ are mutually exclusive. Also, $|A| = |B| + |C| + |D|$. Therefore $$P(A) = {\dfrac{{51\choose 3} \cdot P(3, 2) \cdot 3! \cdot 47! + {51\choose 2} \cdot 3! \cdot P(2, 2) \cdot 47! + {50 \choose 3} \cdot 3! \cdot 3 \cdot 47!} {52!}} \approx 0.002404$$

Is my approach correct? If it is, is there a simpler approach? If not, please guide me to the correct approach via hints/solution. Thank you!
Note
The problem statement previous to this edit was

A Deck of 52 playing cards is shuffled and the cards are turned up and kept on the table. What is the probability that ace of spades and 2 of clubs come right after an ace?

The problem statement turned out to allow for multiple interpretations and @Math Lover's solution is a very elegant one for one of them. I am sorry for the inconvenience caused and will be careful to check that the problem statement is unambiguous before posting it.

Answer 1

Here is how I interpret it -

Question says ace of spades and $2$ of clubs come right after an ace. Number of arrangements where that happens is $2 \cdot 3 \cdot 50!$

Explanation: We need to consider $\{A \ \textbf{A} \ 2 \}$ as one combined card - that is arrangements of $50$ cards. Now the first ace can be any of the $3$ non-spade ace. Also, $2$ of club and ace of spade can swap places as question does not state their internal order.

So the probability should be $ \displaystyle \frac{2 \cdot 3 \cdot 50!}{52!}$