How many tournaments are there? I have absolutely no idea, how can possibly such a problem be solved. Up to isomorphism, how many tournaments are there with five vertices and at least one vertex of in-degree equal to zero? Your answers are much appreciated
 A: I'll consider the two cases of labeled and unlabeled tournaments: in the first case, the vertices are distinguishable, in the second case, tournaments are only distinct up to isomorphism.
There are $2^{\binom n2}$ labeled tournaments on $n$ vertices: for each pair of vertices $(i,j)$, we can choose the orientation of the edge in two ways. If we require one vertex to have in-degree $0$, then there are $n$ ways to choose which vertex that would be, and $2^{\binom{n-1}{2}}$ ways to determine the rest of the tournament, for a total of $n \cdot 2^{\binom{n-1}{2}}$ tournaments. (There can't be two or more vertices with in-degree $0$: the edge between them would give positive in-degree to at least one of them.) In the case $n=5$, we have $5 \cdot 2^6 = 320$ tournaments.
The unlabeled case is harder: the number of unlabeled tournaments on $n$ vertices is given by sequence A000568 in the OEIS, but there's no simple formula like the above. (There is a sum over partitions of $n$...)  We can count the number of unlabeled tournaments on $4$ vertices by brute force, and note that all tournaments on $5$ vertices where one has in-degree $0$ are tournaments on $4$ vertices with an extra vertex $v$ added such that all edges point out of $v$.
To count unlabeled tournaments on $4$ vertices, we can consider what its strongly connected components can look like. 


*

*If the whole tournament is strongly connected, it has a Hamiltonian cycle, and the only tournament we can draw is the first one below.

*We could have strongly connected components of size $3$ and size $1$. There is only one way to choose a strongly connected component of size $3$: a cycle. But then the component of size $1$ can either have all edges pointing into it, or all edges pointing out. This gives the second and third tournaments below.

*There are no strongly connected tournaments of size $2$, so the only remaining case is the transitive tournament, with four strongly connected components of size $1$. This is the last tournament shown below.



A: At first since this is a tournament there can be at most one vertex of zero in-degree. So there is exactly one such vertex. And the question now is how many (up to isomorphism) tournaments are there on four vertices.
The first case is that one vertex has zero in-degree in this subtournament on four vertices. Then we have two different subtournaments on remaining three vertices (one directed cycle and one acyclic digraph).
The second case is that no vertex has zero in-degree and no vertex has zero out-degree in this subtournament on four vertices. Then there are two vertices of in-degree $2$ and two vertices of in-degree $1$. If it easy to see that there is the only such subtournament on four vertices.
And the third case is that no vertex has zero in-degree and one vertex has in-degree $3$ in this subtournament on four vertices. Then there is one more tournament.
Thus we have $4$ tournaments in total.
P. S. Thanks to Misha Lavrov for pointing on the third case.
A: There is another way to count unlabelled tournament
So to count tournaments all you need to do is count the total possible ways in which teams can win or lose.
Here win is (out-degree) and loss is (in-degree) for the given team(vertex)
For example no. of non—cyclic tournaments(unlabelled) for 4 teams say(A,B,C,D)
(I know label does not matter but for the sake of explanation let’s assume a label and then we just don’t count the ones which are isomorphic)
Counting for 3 wins for a team
Case-I
A - 3 wins(wins from B,C and D)
B-  2 wins 1 Loss(wins from C and D)
C-  1 win  2 Losses(wins from D)
D-  3 Losses
Case-II
A - 3 wins(wins from B,C and D)
B - 1 win 2 losses(wins from C)
C - 1 win 2 losses(wins from D)
D - 1 win 2 losses(wins from B)
(One can try and construct more cases for 3 wins but they will all be isomorphic)
(You can also see the image in Misha Lavrov’s answer that only two graphs are there with out-degree 3)
(Interestingly this is like a sub-graph kind of thing. As if you ignore A in both the cases then you get no. of tournaments if n=3(though one is cyclic and other is acyclic)
Counting for 2 wins for a team
Case-1
A- 2 wins 1 Loss(won from B,C)
B- 2 wins 1 Loss(won from C,D)
C- 1 win  2 Losses(won from D)
D- 1 win  2 losses(won from A)
Case-2
A - 2 Wins 1 Loss(won from B and D)
B - 2 Wins 1 Loss(won from C and D)
C - 2 Wins 1 Loss(won from A and D)
D - 3 Losses
Now since you can’t build a case of 3 wins(already counted) and if A  then it becomes mandatory that there exist at least one more team(say B) who win two games and hence can be seen these are the possible two cases.
So total cases - 4 (Phew!)
