GRE permuation and combination problem 1

Question

Question: The University of Maryland, University of Vermont, and Emory University have each $4$ soccer players. If a team of $9$ is to be formed with an equal number of players from each university, how many possible teams are there?

Answer: The selection from the $3$ universities can be done in $4\times 4\times 4= 4^{3}=64$ ways.

I don't understand this solution. How come a team of $9$ be formed like that? Maybe I don't understand the question entirely.

Answer 1

Since they need to have an equal number of players from each university, it means it needs to have $3$ players from each university.

If each university has $4$ different players, how many different subsets (combinations) of 3 players can you get from a total set of $4$ players? Obviously $\binom{4}{3}=4$

So $4$ different valid combinations that each university can provide, make a total of $4^3=64$ total possibilities.

Answer 2

Each university has 4 players and you have to choose 3 players from each university.

$$\binom{4}{3}\times\binom{4}{3}\times\binom{4}{3}=4\times 4\times 4= 4^{3}=64$$