Additional Info: 24 students in a Zoom Meeting that has 4 breakout rooms. Each breakout room has 6 students. Additionally there are 6 students belonging to each of the 4 grade levels. I want the probability that at least one breakout room has students all from the same grade.
$$ \text{Your expression (part of a larger work): } \binom{4}{1} \cdot \left(\frac{\binom{6}{6}}{\binom{24}{6}}\right) - \binom{4}{2} \cdot \left(\frac{\binom{6}{6}}{\binom{24}{6}} \cdot \frac{\binom{6}{6}}{\binom{18}{6}}\right) \ldots $$
I know the number of terms in each sum in the Inclusion-Exclusion Principle are represented by the 4C1,4C2 and my thinking is that for the case of one room having all students in the same grade year that you take into account all the ways to choose 6 from 24 which represents all the ways students can be placed into rooms, and then 6C6 to represent choosing 6 students from one grade year. I think I am missing something and don't believe this is right
Let $E_1$ be the event that the first grade level is entirely together, and let $E_2,E_3,E_4$ be defined analogously. First of all, $$ P(E_1)=4\times \frac{\binom 66}{\binom{24}6}=\frac{5\times 4\times 3\times 2\times 1}{23\times 22\times 21\times 20\times 19} $$ Explanation 1: Label the people in the first grade level A through F.
The probability that B is in the same room as A is $\frac{5}{23}$. This is because there are five locations where B could be in the same room as A (the five other spots on A's team), and there are 23 locations where B could be total (anywhere but where A is).
Conditional on B being in the same room as A, the probability that C is in the same room as A and B is $\frac 4{22}$.
Continuing in this fashion, you get $\frac{5\times 4\times 3\times 2\times 1}{23\times 22\times 21\times 20\times 19}$.
Explanation 2: Number the breakout rooms $1,2,3,$ and $4$. You claimed that $ {\binom 66}/{\binom{24}6}$ was the probability that grade level one was all together, but in fact, it is more specfically the probability that they are all together in a particular breakout room. To get the probability they are altogether in any breakout room, you must multiply by four.
Using the same logic, you can show that your second term is off by a factor of $4\times 3$ as well. That is, $$ P(E_1\cap E_2)=4\times 3\times \frac{\binom 66}{\binom {24}6}\cdot \frac{\binom66}{\binom{18}6} $$ Therefore, the whole expression should begin $$ \binom{4}{1} \cdot \left(\color{green}4\times \frac{\binom{6}{6}}{\binom{24}{6}}\right) - \binom{4}{2} \cdot \left(\color{green}{4\times 3}\times\frac{\binom{6}{6}}{\binom{24}{6}} \cdot \frac{\binom{6}{6}}{\binom{18}{6}}\right)+ \ldots $$