 Solution

Count unordered sets of distinct labelled fish: there are $\left(\genfrac{}{}{0ex}{}{N}{M}\right)$ possible hauls. To obtain exactly $j$ trout, choose $j$ labels from the $K$ trout and $M-j$ from the $N-K$ carp, giving $\left(\genfrac{}{}{0ex}{}{K}{j}\right)\left(\genfrac{}{}{0ex}{}{N-K}{M-j}\right)$ possibilities. Uniform sampling without replacement therefore gives the hypergeometric distributionOutside this integer range the probability is zero. If ordered catches were counted instead, every haul would have $M!$ orders and that factor would cancel from the probability ratio.

Here equal catchability means equal selection probability for each individual remaining fish. It does not mean that the two species have probability $1/2$ on each draw regardless of their populations. This interpretation is the one that yields the stated hypergeometric law.