Question: A school has 4 classes. Each class consists of 25 students. The school needs to choose 10 class representatives, in which there is a minimum of one student from each class. How many ways there are to do so?
I get why the final answer, using inclusion/exclusion, will be: $${{100}\choose{10}} - {{4}\choose{3}}{{75}\choose{10}} + {{4}\choose{2}}{{50}\choose{10}} - {{4}\choose{1}}{{25}\choose{10}}$$
But, why couldn't I do something like the following: First, I need to assure that there 4 students, each representing a separate class. In total, there are $25^{4}$ ways of doing so. After that, I can choose the 6 available slots for the class representatives from the $100-4 = 96$ students left, meaning that there are ${96}\choose{6}$ ways of doing so. In total, from what I just described, where would be a total of: $${25^{4}}*{{96}\choose{6}}$$ possibilities for the selection.
Conceptually I must be getting something wrong. What am I over-counting/under-counting?
That seems to be a over-counting. Say students in classes $A$, $B$, $C$, $D$ are named such as $a_1,...,a_{25}$ for class $A$; $b_1,...,b_{25}$ for class $B$; $c_1,...,c_{25}$ for class $C$, $d_1,...,d_{25}$ for class $D$. Then when we choose one student from each class first, assume we have chosen $a_1$, $b_1$, $c_1$ and $d_1$. Then assume, rest of the students are $a_2$, $a_3$, $a_4$, $a_5$, $a_6$ and $a_7$. But, this is as same as chosing $a_2$ first instead of $a_1$, then choosing $a_1$, $a_3$, $a_4$, $a_5$, $a_6$ and $a_7$. So we are overcounting.
NOTE: You can easily see that the example I gave is not the only case we are counting more than once.