Here's the statement problem:
"From an ordinary deck of playing cards, cards are to be drawn successively at random and without replacement. The probability that the third spade appears on the sixth draw is computed as follows: Let A be the event of two spades in the first five cards drawn, and let B be the event of a spade on the sixth draw. Thus, the probability that we wish to compute is P(A ∩ B)."
The conditional probability approach: We have that $\frac{\left( \begin{array}{c} 13 \\ 2 \end{array} \right) \left( \begin{array}{c} 39 \\ 3 \end{array} \right)}{\left( \begin{array}{c} 52 \\ 5 \end{array} \right)}$ and $P(B|A) = \frac{11}{47}$, so the desired probability, $P(A \cap B) = P(A)P(B|A)= \frac{\left( \begin{array}{c} 13 \\ 2 \end{array} \right) \left( \begin{array}{c} 39 \\ 3 \end{array} \right) \times 11}{\left( \begin{array}{c} 52 \\ 5 \end{array} \right) \times47}$
The classical approach We have that $P(A \cap B) = \frac{N(A\cap B)}{N(S)}$, where $S$ is our sample space. We have that $N(A\cap B) = \left( \begin{array}{c} 13 \\ 2 \end{array} \right) \left( \begin{array}{c} 39 \\ 3 \end{array} \right){\left( \begin{array}{c} 11 \\ 1 \end{array} \right)}$, because we want to choose 2 spades out of 13 spades, then 3 non-spades of the remaining 39 cards, and then 6th position is one the spades in the 11 spades. On the other hand, $N(S) = \left( \begin{array}{c} 52 \\ 6 \end{array} \right)$, so we have $P(A \cap B) = \frac{\left( \begin{array}{c} 13 \\ 2 \end{array} \right) \left( \begin{array}{c} 39 \\ 3 \end{array} \right){\left( \begin{array}{c} 11 \\ 1 \end{array} \right)}}{\left( \begin{array}{c} 52 \\ 6 \end{array} \right)}$.
The two approaches should yield the same result, but here they' re not. Also, I noticed that while the numerators are the same, the denominators are different (though not too much). Where was I wrong?
The first calculation is the correct one.
To convince you of this, you can take your attempts and try to apply the same formula to smaller examples. Briefly take a look at the related question of where we have a three-card deck: an ace of spades, a two of spades, and an ace of hearts. We ask the question of what is the probability that we draw our second spade as the third card.
By brute force listing out of the six possibilities, you should see that four of the six yield a spade as the last card.
Using the first approach, we would arrive at the same correct answer: $\frac{\binom{2}{1}\binom{1}{1}}{\binom{3}{2}}\cdot\frac{1}{1}=\frac{2}{3}$
Using the second approach, we would have arrived at a wrong answer: $\frac{\binom{2}{1}\binom{1}{1}\binom{1}{1}}{\binom{3}{3}}=2$, which happens to be off by a factor of three.
To fix your second calculation, it might be easiest to explain by instead of working in the scenario that order of cards in hand don't matter, that we work in the scenario where order of cards in hand do matter.
We approach the same way: $\binom{13}{2}$ ways to pick the early spades, $\binom{39}{3}$ ways to pick the other early cards, and then $\binom{11}{1}$ ways to pick the final spade. We then pick in what specific order the cards appeared in $5!$ ways, noting that the final spade must occur in the final position and nowhere else.
The bottom, similarly, will be $\binom{52}{6}6!$ as after picking what six cards apppear, we can arrange them in $6!$ ways.
This gives our corrected calculation at $$\frac{\binom{13}{2}\binom{39}{3}\binom{11}{1}5!}{\binom{52}{6}6!}=\frac{1}{6}\cdot\frac{\binom{13}{2}\binom{39}{3}\binom{11}{1}}{\binom{52}{6}}$$
which one can verify is equal to the answer given by your first attempt.
As discussed in comments elsewhere, that $6$ happens to be equal to $3!$ is merely a coincidence. We divide by $6$ since that is the number of cards in the hand, just as dividing by three and not by $2!$ corrected the calculation in the smaller three-card deck example above.
A different way of picturing the correction is that in the second attempt without the fix, we did not account for the final card actually appearing in the final position. The final card actually appears in the final position with probability $\frac{1}{6}$ (or more generally $\frac{1}{n}$ if referring to drawing $n$ cards), and so this should have been multiplied in as well.
Another example to try to convince that dividing by $n$ is the correct fix and not dividing by the factorial of the number of spades:
Consider the problem of asking what the probability that the first spade occurs in the $6$'th draw.
The first approach, again the correct one, would give an answer of $\frac{\binom{39}{5}}{\binom{52}{5}}\cdot\frac{13}{47}$
The second approach without the fix would give an answer of $\frac{\binom{39}{5}\binom{13}{1}}{\binom{52}{6}}$, an answer six times larger than the first, just as before, despite the number of spades having changed between the examples. Dividing by the factorial of the number of spades would not have corrected it.