Finding a conditional probability for draws from an urn

An urn contains 10 red, 6 blue, and 8 black marbles. What is the probability that 3 black marbles are simultaneously selected given that 1 blue marble was already selected?

I am obtaining two different answers based on my approach. I guess that approach 2 is closer to the correct answer, but I cannot see the flaw in the reasoning.

Approach 1: Since 1 blue marble was selected (and not replaced), there remain 23 marbles of which 8 are black, 10 are red, and 5 are blue. Therefore, the probability of now choosing 3 black marbles simultaneously is

$$\frac{{8 \choose 3}}{{23 \choose 3}}=0.0316.$$

Approach 2: Using the conditional probability formula $$P(A\mid B)=\frac{P(A\&B)}{P(B)},$$ we obtain

$$\frac{\frac{{6 \choose 1}{8 \choose 3}}{{24 \choose 4}}}{\frac{{6 \choose 1}}{{24 \choose 1}}}=0.126.$$

The correct probability is the first one: Given that the first marble drawn is blue and not replaced, the probability that the next three marbles drawn without replacement are all black is $\binom{8}{3}/\binom{23}{3} = \frac{8}{253}$.

Why is the second computation wrong? The answer becomes clear if we examine the joint probability $\Pr[A \cap B]$ more closely. You wrote $$\Pr[A \cap B] = \frac{\color{blue}{\binom{6}{1}}\binom{8}{3}\color{red}{\binom{10}{0}}}{\binom{24}{4}},$$ where I have added the factor $\binom{10}{0}$ to illustrate that this is a generalized hypergeometric probability we are calculating. But this is not correct, because it ignores the sequential ordering of the outcomes implied by the observation of the events $A$ and $B$. In particular, the event $A \cap B$ includes the following outcomes: $$(blue, black, black, black) \\ (black, blue, black, black) \\ (black, black, blue, black) \\ (black, black, black, blue)$$ In other words, $A \cap B$ allows the outcome in which the first draw was black, then exactly two black and one blue marble is drawn from the second draw of three marbles, because the hypergeometric distribution models the drawing process as if all marbles are taken without respect to ordering. This is disallowed under the original interpretation of the question.

You will also note that this difference is reinforced when we see that under the second computation, the conditional probability calculated is $4$ times the first calculation.