I have 3 problems about this combinational problem.
I have 24 students.
1. Firstly, In how many ways we can assign 24 students in 3 groups if the first group has 14 students, second group 6 and third group 4?
My Αnswer:
$$ \binom{24}{14} \cdot \binom{10}{6} \cdot \binom{4}{4} = 411863760. $$
2. Secondly, In how many ways we can assign 24 students in 3 groups if each group has 8 students?
My Αnswer:
$$ \frac{\binom{24}{8} \cdot \binom{16}{8} \cdot \binom{8}{8}}{3!} = 1577585295. $$
3. Thirdly, In how many ways we can assign 24 students in 3 groups if the first group has 14 students, second group 5 and third group 5?
My Αnswer:
$$ \binom{24}{14} \cdot \frac{\binom{10}{5} \cdot \binom{5}{5}}{2!} = 247118256. $$
4. Finally, In how many ways we can assign 24 students in 3 groups if the first group has 14 students, second group 5 and third group 5 BUT the first group is named RED, the second GREEN and the third BLUE?
My Αnswer:
$$ \binom{24}{14} \cdot \binom{10}{5} \cdot \binom{5}{5} = 494236512. $$