10 fish are caught. (Probability Question) Question says: Suppose that 10 fish are caught at a lake that contains 5 distinct types of fish.
a) how many different outcomes are possible, where an outcome specifies the numbers of caught fish of each of the 5 types?
(okay so I am not 100% clearly understood after reading the question. So suppose there are like 2 A, 2 B, 2 C, 2 D, 2 E fish. The outcome is 2 of each 5 types of fish. So is it like 5x5x5x5x5x5x5x5x5x5, 5^10? but then again what if some types are not caught?? It sounds so simple but it's confusing me a lot. Any tips would be helpful.)
b) how many outcomes are possible when 3 of the 10 fish caught are trout?
(so three of the 10 fish have to be one type. That means 4x4x4x4x4x4x4x(3x2x1)? is this correct?)
Thank you!
 A: a) One possible outcome is (2,2,2,2,2), which indicates 2 of type A, 2 of type B, etc.  Another possible outcome is (9,0,1,0,0), which indicates 9 of type A, 1 of type C.  Hence you have an ordered list of five nonnegative integers, whose sum is 10.  Count these.
b) Now we know that the first element of the list is 3, so we have four remaining digits, that sum to 7.  For example (3,7,0,0,0) or (3,2,2,2,1).  Hence you can count ordered lists of four nonnegative integers, whose sum is 7.  Count these.
A: I am unsure whether my answer is correct or not. Someone please comment. But first let me put the full question as it is from its source.

Suppose that 10 fish are caught at a lake that contains 5 distinct types of fish.
(a) How many different outcomes are possible, where an outcome specifies the numbers of caught fish of each of the 5 types
(b) How may outcomes are possible when 3 of the 10 fish caught are trout?
(c) How many when at least 2 of the 10 are trout?

Let me add one more to above (please comment if the solution I wrote for this one is correct or not)

(d) What will be the solution to (c) when 5 different types of fishes are caught?

Solution
(a) This is star and bar theorem 2 problem. Let us assume there are 10 stars (representing fishes) put in a row. Now we separate them into 5 groups (representing fish types) by putting 4 bars among them. (Notice that if we put bar before first star on extreme left, it will denote that the first group is empty. Same for bar put after last star and two bars put together without any star between them) So we have total 14 positions (10 stars and 4 bars) out of which we have to select 4 for bars which can be done in $\binom{10+5-1}{5-1}=\binom{14}{4}=1001$ ways.
(b) When 3 out of 10 are trout, there can be maximum 4 different types of fishes in remaining 7. With same above logic, this can be done in $\binom{7+4-1}{4-1}=\binom{10}{3}=120$ ways.
(c) When at least 2 out of 10 are trout, there can be maximum 5 different types of fishes in remaining 8. This can be done in $\binom{8+5-1}{5-1}=\binom{12}{4}=495$ ways
(d) This becomes star and bar theorem 1 problem. When there is exactly 2 trouts, there should be 4 types of fishes in remaining 8. So not we order 8 stars in a row. They will have 7 gaps separating them. We put 3 bars in those gaps to separate them into 4 groups. This can be done in $\binom{8-1}{4-1}=\binom{7}{3}=35$ ways. If there are more than 2 trouts, then the remaining 8 fishes must contain 5 types of fishes. This can be done in $\binom{8-1}{5-1}=\binom{8-1}{5-1}=\binom{7}{4}=35$ ways. By product rule we get $35\times 35=1225$
