Using combinations to combinations for probability with dependent events

Question

I am reading a problem about probabilities with dependent events.

There are $3$ people selected from $5$ freshmen, $2$ sophomores and $4$ juniors. What is the probability that the first $2$ people selected are freshmen and the third a junior?

I thought that this is solvable by using combinations.
I.e.
There are $\binom{11}{3}$ ways to select $3$ out of $11$ students (I hope that the notation is for combinations and not permutations as I get these 2 notations mixed).
This is $$\frac{11!}{3!\cdot 8!} = 165$$ ways
To choose $2$ freshman from $5$ is: $$\frac{5!}{2!\cdot 3!}$$ To choose $1$ junior from $4$ is: $$\frac{4!}{1!\cdot 3!}$$
So the probability we are looking for is:
$$\frac{\frac{5!}{2!\cdot 3!} \cdot \frac{4!}{1!\cdot 3!}}{165} = \frac{40}{165}=\frac{8}{33}$$

but the answer is wrong.
Is it not possible to use the combinations formula for calculating such a probability with $3$ dependent events?

Answer 1

Your answer would be correct if the question were:

What is the probability that two of the people selected are freshman and one is a junior?

That would be a question about combinations - how many ways to choose 2 freshman out of 5 and one junior out of 4, over the number of ways to choose a group of 3 students from 11.

That is subtly different from the question which was asked. The question in your post says the first two people are freshman, and the third is a junior - i.e., the order of selection matters. So you should instead use permutations.

Answer 2

When a particular order is specified, it is simplest to directly multiply probabilities,

so $\dfrac5{11}\cdot\dfrac4{10}\cdot\dfrac4{9}$

If you want to use combinations, you will have to take an extra step which unnecessarily lengthens the process, viz divide by $\dbinom32$,

thus $\dfrac{\dbinom52\dbinom41}{\dbinom{11}3}\div\dbinom32$