$75$ students in three balanced groups: Probability, cardinality. EDIT: more consistent attempt.
Let say we have $n=75$ students, divided into 3 groups.

What is the probability that the three groups are balanced and the probability that $Marc$ and $Gato$ are in the same group?

Let $\Omega=\{(m_1,m_2,m_3)\in\{0,\cdots,75\}^3: m_1+m_2+m_3=75\}$
Let $\omega_1\in\Omega : \omega_1=(m_1,m_2,m_3)$ representes the number of studies in each group. Question: How can I compute card$(\Omega)$?
Another approach: 
$\Omega_2=\{(E_1,E_2,E_3), E_i\in \Gamma : E_1\cup E_2\cup E_3, \quad E_i\cap Ej=\emptyset \}$
$$\text{Card}(\Omega_2)=\sum_{k_1+k_2+k_3=75}\dbinom{75}{k_1}\dbinom{75-k_1}{k_2}\dbinom{75-k_1-k_2}{k_3}=3^{75}.$$
How can I continue? 
 A: There are 3 possible groups, and 75 people in total, so the total possibilities are: $3^{75}.$
If we choose a group for Marc and Gato, we can do so in 3 ways. Now, we have a group with only 2 people in it, so to fill the rest of the group, we chose 23 people from the 72 remaining. To make the second group, we choose 25 from the leftover 50. 
So I believe the probability is something like:$$\frac{3\cdot\binom{73}{23}\binom{50}{25}}{3^{75}}$$
However, I'm not sure whether we'll need to divide by a factor of 2 or not as the remaining two groups may be indistinguishable. 
A: The OP do not specify very clearly how the groups are decided, I suppose here
that each student chooses randomly one of the three groups with equiprobability.
The natural model is then $\Omega=G^S=\lbrace f : S\to G \rbrace$ where
$G=\lbrace G_1,G_2,G_3 \rbrace$ is the set of groups and
$S=\lbrace S_1,S_2,\ldots, S_{75} \rbrace$ is the set of all students. Note that
$|\Omega|=3^{75}$.
The probability $p_1$ that the groups are balanced is
$$
p_1=\frac{\binom{75}{25}\times\binom{75-25}{25}}{3^{75}}=\frac{27356996443592349250523576343392}{2503155504993241601315571986085849} \approx 0.01
$$
If we put $S_1=Marc$ and $S_2=Gato$, the probability $p_2$ that Marc and Gato belong to the same group is the probability that $f(2)=f(1)$ for $f\in \Omega$, i.e. $\frac{3}{9}=\frac{1}{3}$.
