So we can think of the multinomial distribution like the binomial distribution except that there are more than two outcomes.
In the binomial distribution, each trial is independent of each other(i.e. Bernoulli Trials), so we use the binomial distribution in probabilistic situations where we draw with replacement. Likewise, we should also be using the multinomial distribution for situations where we draw with replacement, correct?
However, the following solution doesn't hold if the above is True. For reference, this comes from "The Probability Tutoring Book" by Carol Ash.
60% of students are against a policy, 30% of students are for the policy and 10% of students are indifferent. Find the probability that of ten students selected at random, 4 are for the policy, 5 are against the policy, and 1 is indifferent.
The answer
$\frac{10!}{4! 5! 1!}(.6)^5(.3)^4(1)^1$
I understand where the fraction comes from - we can arrange the 10 students in $10!$ ways, choosing the same 4 students just in a different ordering is double counting them so we divide by $4!$ and likewise for the students against the policy.
However, isn't the multinomial distribution only applicable if we draw with replacement, in this case, we can't choose a person twice when we select the 10 students, so why are we allowed to do so here?