I'm supposed to solve this for my homework, but i'm facing some problems so i hope to receive some advise.
Suppose that 10 fish are caught at a lake that contains 5 distinct types of fish. How many different outcomes are possible, where an outcome specifies the numbers of caught fish of each of the five types?
Is it possible that i've caught zero of certain type of fish? Since i have 10 tries and 5 types to choose from each time, do i just use 5^10?
Is it possible that I've caught zero of certain type of fish?
Yes. Indeed you could have caught $10$ fish of same kind. This is called a variation with repetition or $V_{{R^m_n}}$ where:
$m$ is the number of different kind of fish.
$n$ is the number of total caught fish.
The formula is $m^n$. Since each caught fish can be of five types you have $5^{10}$ so your guess is right.
Pay attention to the phrase: an outcome specifies the numbers of caught fish of each of the five types. The value $5^{10}$ calculates the total amount of permutations with repetition. This value supposes that the outcomes $$(1,1,1,1,1,1,1,1,1,2) \text{ and } (2,1,1,1,1,1,1,1,1,1)$$ are different since the order here differ.
And according to the condition of the problem, they are the same. In both cases the first kind of fish is caught 9 times, and the second one is caught once, so it is the same outcome.
Therefore one should use combinations with repetition to calculate how many different outcomes are possible $$ \binom{n+m-1}{n}=\binom{10+5-1}{10}. $$ This value calculates the total number of outcomes, if one outcome is described by the numbers of caught fish of each of the five types. Say, here are possible outcomes: $(10,0,0,0,0)$ - 10 fish of first kind, no other fish, $(9,0,1,0,0)$ - 9 fish of first kind, one fish of third kind, $(0,2,0,0,8)$ - 2 fish of second kind, 8 fish of last kind ans so on.
Not quite.
There are indeed $5^{10}$ ways to draw ten fish of the five types if order is a distinction. However the problem specified that an outcome is to be distinguished only by counts of fish of each type; so order is not a distinction.
That is that three trout then one catfish followed by a cod is the same outcome as a cod then three trout followed by a catfish, or any of the other $5!/3!$ arrangements. Thus there are far fewer distinct outcomes when order is not a distinction.
When order is not a distinction the count of ways to select ten fish of the five types is the same as the count for placing ten indistinguishable balls into five distinct boxes. That is :
$$\binom{14}{10}$$