What is the probability that one team is not represented in the five selected players? A total of five players is selected at random from four sporting teams. Each of the team consists of ten players numbered from $~1~$ to $~10~$
What is the probability that one team is not represented in the five selected players?
This is my working out to the question 
$$P = \frac{(^{10}C_1)^{3}\times ^{27}C_2}{^{40}C_5}  = \frac{375}{703} $$
However, it is wrong and here is my teacher's solution
$$P = \frac{(^{10}C_3)\times (^{10}C_1)^{2} \times 3\times 4 + (^{10}C_2)^{2} \times ^{10}C_1 \times 3\times 4}{^{40}C_5}  =  \frac{5375}{9139} $$
I noticed that my value is smaller than his but I couldn't find where I got wrong. I actually think that my teacher's answer is incorrect since regarding the numerator, there might be some cases in which he considered 3 players from a team and one from each of the other two overlapping the second case when he considered 2 players from the first team and two from the second team and 1 from the other team. 
However, I really don't know. Please help me to figure out which part in my solution (or his) is wrong.
Thank you so much in advance!
 A: For example, if there're four teams A,B,C,D.
There're 4 different ways that one team is not present, such as team A is not present, or team B is not present.
So in your formuala: $(^{10}C_1)^{3}\times ^{27}C_2$
i) It doesn't count which team is not present. 
ii) $(^{10}C_1)^{3}$ means select one member from three teams. In the next step $^{27}C_2$ means selecting any two members from the three teams but some choices are counted multiple times. 
For example if in the first step, Alice from Team A, Bob from Team B and Cindy from Team C are selected and in the next step, Adam and Anwar from Team A are selected so we reach a choice (Alice, Adam, Anwar, Bob, Cindy).
But if in the first step, Adam from Team A, Bob from Team B and Cindy from Team C are selected and in the next step, Alice and Anwar from Team A are selected, we could reach the same choice (Alice, Adam, Anwar, Bob, Cindy) too.
The teacher's solution should be right since to select 5 members frmo exact 3 team, either one team contributes 3 members or two teams both contribute 2 members.
The $3\times 4$ in $(10C3)×(10C1)2×3×4$ is used to determine which team contributes 3 members and which team is not present. 
A: Let $m_i$ be the event that we miss a player from team $i$.
We can apply the inclusion-exclusion formula to find the probability we miss some team:
$$P(m_1 \cup m_2 \cup m_3 \cup m_4 ) = \sum_{i=1}^4 P(m_i) - \sum_{(i,j)} P(m_i \cap m_j) + \\
+ \sum_{(i,j,k)} P(m_i \cap m_j \cap m_k) $$
where we sum over all distinct $k$-tuples of team numbers. We stop at $3$-tuples, as we cannot miss all the teams. 
Now, $$P(m_i)= \frac{\binom{30}{5}}{\binom{40}{5}}$$ (we pick 5 players from only 40 possible ones: all players minus the $10$ from team $i$) and this holds for all $5$  teams. So (thanks to the symmetry) the first sum is $4\frac{\binom{30}{5}}{\binom{40}{5}}$
$$P(m_i \cap m_j) = \frac{\binom{20}{5}}{\binom{40}{5}}$$ (we omit 20 players) and this holds for $\binom{4}{2}=6$ many $2$-tuples of teams. So the second sum is $6\frac{\binom{20}{5}}{\binom{40}{5}}$
Similarly you miss three fixed teams we are left with a probability
of $$P(m_i \cap m_j \cap j)= \frac{\binom{10}{5}}{\binom{40}{5}}$$
We have $\binom{4}{3}=4$ many $3$-tuples of teams to omit. (Or just 4 teams to use exclusively). So the third and final sum
equals $4\frac{\binom{10}{5}}{\binom{40}{5}}$ 
Now take the alternating sum of those. Wolfram alpha tells us it equals $$\frac{6639}{9139}$$ which is about $0.726$
So my answer disagrees with both of you, I'm afraid. As N.F. Taussig points out in the comments, this is due to a difference in interpretation of "one team is not represented". I assume that this means that at least one team is not represented, possibly more, while the teacher's solution assumes exactly one team is not represented. As the question is possibly a translated one, I will not venture into a discussion which interpretation is the correct one in this case.
