Reduced Sample Space solution different than Conditional Probability Consider the following problem,
The theaters of a town are showing seven comedies and nine dramas. Marlon has
seen five of the movies. If the first three movies he has seen are dramas, what is
the probability that the last two are comedies? Assume that Marlon chooses the
shows at random and sees each movie at most once.
The solution manual solves it as follows:
Reduce the sample space: Marlon chooses from six dramas and seven comedies two at random.
What is the probability that they are both comedies? The answer is $C(7, 2) / C(13, 2) = 0.2692307692307692$.
Now let's solve it by using conditional probability formula itself, without reducing the sample space. Let $A$ be the event that Marlon chooses the first three dramas, and $B$ the event that the other two are comedies. Thus we have,
$$
P(B | A) = \frac{P(A\cap B)}{P(A)} = \frac{\frac{C(9, 3)\times C(7, 2)}{C(16, 5)}}{\frac{C(9, 3)\times C(7, 2) + C(9, 4) \times C(7, 1) + C(9, 5) \times C(7, 0)}{C(16, 5)}} = 0.6363636363636364
$$
So why are the answers different?
 A: The error with your approach is that although the problem statement talks about "first movie seen" and "second movie seen", your sample space did not make mention of what the order you saw the movies in was.  You can correct this by keeping track of the entire order in which you saw the movies... so rather than $\binom{16}{5}$ options for the movies seen in all, you instead have $16\times 15\times 14\times 13\times 12$ options.  Another option is that you can imagine the "first three movies" seen all happening in a clump and the final two movies happening in a clump together... giving $\binom{16}{3}\binom{13}{2}$ options.  Your approach seems to have tried to use this second idea, however you were not consistent with how the numerator and denominator treats these clumps as being notably different.  Alternatively, you incorrectly counted some occurrences of having watched three dramas as being scenarios where "the first three movies seen were dramas" and you had included things such as where you started with two dramas and a comedy in the first clump and saw a drama and comedy in the second clump which should not have counted.
The order in which things happen and whether we solve the problem with order mattering vs order not mattering will in most problems not matter... however we must be consistent with whether order does or does not matter for numerator and denominator, and we must be able to accurately and unambiguously describe the desired events with our selection.
With full order mattering:
$$\Pr(A\mid B) = \dfrac{\Pr(A\cap B)}{\Pr(B)} = \dfrac{~~~\frac{9\cdot 8\cdot 7\cdot 7\cdot 6}{16\cdot 15\cdot 14\cdot 13\cdot 12}~~~}{~~~\frac{9\cdot 8\cdot 7 \cdot 13\cdot 12}{16\cdot 15\cdot 14\cdot 13\cdot 12}~~~}$$
With only "firstclump" vs "secondclump" mattering:
$$\Pr(A\mid B) = \dfrac{\Pr(A\cap B)}{\Pr(B)} = \dfrac{~~~\frac{\binom{9}{3}\binom{7}{2}}{\binom{16}{3}\binom{13}{2}}~~~}{~~~\frac{\binom{9}{3}\binom{13}{2}}{\binom{16}{3}\binom{13}{2}}~~~}$$
Both of course simplify to $\dfrac{7}{26}$ as expected
