Number of ways of selecting a mathematics team and an informatics team with conditions If we have $45$ different students ($20$ students from 12th grade, $25$ students from 11th grade). If we want to choose $15$ students for a mathematics team and $10$ students for am informatics team. 
1) How many ways can we achieve this if the mathematics team was to have only 12th grade students? 
2) How many ways can we do this if mathematics team has to have at least $1$ student from 12th grade?
I have thought of solving this like this: 
1) $C(20,15) \cdot C(30,10)$ (We have to choose $15$ from the $20$ of the 12th grade and then $10$ from the thirty remaining we don't care what grade they are)
2) Same logic: $C(20,1) \cdot C(44,10) \cdot C(30,10)$
$C$ is combination and $C(a,b)$ means $$\frac{a!}{(a-b)!b!} $$
Is my solution correct? I am not sure because I haven't quite understood the difference between permutations and combinations when applied to this kind of problems.
 A: The second answer is wrong, because in the $C(44,10)$ (which should be $C(44,14)$, noted $\binom{44}{14}$), there could be initial choice of the student in 12th grade already counted (not clear, I know, english is not my first language).
Let me explain : you take combinations because you are NOT interested by order : let's say students from 12th grade are named A,B,...,T. You chose A to have one student from 12th grade, then 14 students to complete the math team, INCLUDING B. This case is also counted when you start by choosing B to have one 12th grade, THEN choose 14 other students, INCLUDING A. See the problem ?
When you encounter events of the form "at least one...", always consider the opposite event, it is generally simpler. 
For example here : there are $\binom{45}{15}.\binom{30}{10}$ different teams compositions. The number of choices where the math team does NOT include a 12th grade student is $\binom{20}{15}.\binom{30}{10}$. Therefore the correct answer for second question is 
$$\binom{45}{15}.\binom{30}{10}-\binom{20}{15}.\binom{30}{10}$$
A: If no student may be a member of both teams, your answer to the first question is correct.  However, your answer to the second question is incorrect.

In how many ways can a mathematics team with $15$ students and an informatics team with $10$ students be selected from $45$ students composed of $20$ 12th grade students and $25$ 11th grade students if the mathematics team must have at least one $12$th grade student?

Method 1:  If we assume no student there were no restrictions, we could choose the mathematics and informatics teams in 
$$\binom{45}{15}\binom{30}{10}$$
where we select $15$ of the $45$ students to be on the mathematics team and $10$ of the remaining $45 - 15 = 30$ students to be on the informatics team.
We must subtract those selections in which none of the students on the mathematics team are in the 12th grade.  If no 12th grade students are selected for the mathematics team, then we must $15$ of the $25$ 11th grade students for the mathematics team and $10$ of the remaining $30$ students for the informatics team, which can be done in 
$$\binom{25}{15}\binom{30}{10}$$
ways.  
Hence, the number of admissible selections is
$$\binom{45}{15}\binom{30}{10} - \binom{25}{15}\binom{30}{10}$$ 
Method 2:  The number of ways of selecting exactly $k$ 12th grade and $15 - k$ 11th grade students for the mathematics team is 
$$\binom{20}{k}\binom{25}{15 - k}$$
Since at least one 12th grade student must be on the mathematics team, the number of ways of selecting the mathematics team is 
$$\sum_{k = 1}^{15} \binom{20}{k}\binom{25}{15 - k} = \binom{20}{1}\binom{25}{14} + \binom{20}{2}\binom{25}{13} + \binom{20}{3}\binom{25}{12} + \cdots + \binom{20}{15}\binom{25}{0}$$
Therefore, the number of ways of selecting a mathematics team and an informatics team if at least one 12th grade student must be on the mathematics team is
$$\binom{30}{10}\sum_{k = 1}^{15} \binom{20}{k}\binom{25}{15 - k}$$
Why is your answer incorrect?
I assume you meant to write 
$$\binom{20}{1}\binom{44}{14}\binom{30}{10}$$
since there are $14$ additional members of the mathematics team to be selected.  However, this would still be incorrect.
By designating a particular 12th grade student as the 12th grade student on the mathematics team, you have counted each selection in which there are $k$ 12th grade students on the mathematics team $k$ times, once for each way of designating one of those students as the 12th grade student on the mathematics team.  Therefore, your count
$$\binom{20}{1}\binom{44}{14}\binom{30}{10} = \binom{30}{10}\sum_{k = 1}^{15} \color{red}{\binom{k}{1}}\binom{20}{k}\binom{25}{15 - k}$$
