Forming a committee from groups with conditions There are three groups of 10 individuals - 30 individuals in total.
From these, choose 10 individuals, including at least 2 from each group.
It seems to me that the answer to this should be $C(10,2)*C(10,2)*C(10,2)*C(24,4)$,
where C(10,2) is the choice of 2 individuals from each group, and $C(24,4)$ is the choice of 4 individuals from the remaining $24$.
Apparently, this is not correct.
Where am I going wrong?
 A: Directly:
$$\sum_{i=2}^6 \sum_{j=2}^{8 - i} \binom{10}{i} \binom{10}{j} \binom{10}{10-i-j} = 24466050$$

Via inclusion-exclusion, where the three properties to be avoided are that group $g$ has fewer than two individuals selected:
\begin{align}
&\binom{30}{10}-\binom{3}{1}\left[\binom{10}{0}\binom{20}{10}+\binom{10}{1}\binom{20}{9}\right]+\binom{3}{2}\left[\binom{10}{0}^2\binom{10}{10}+\binom{2}{1}\binom{10}{0}\binom{10}{1}\binom{10}{9}+\binom{10}{1}^2 \binom{10}{8}\right] \\
&= 30045015 - 5593068 + 14103 \\
&= 24466050\end{align}
A: "What is my mistake ?" : You are making overcounting.Your combination calculating is prone to count one individual more than once
You can get rid of overcounting by using generating functions such that
If there will be at least $2$ people from each group , then a group can give at most $6$ people.
The generating function of a group is $$(C(10,2)x^2 +C(10,3)x^3+C(10,4)x^4+C(10,5)x^5+C(10,6)x^6)$$
Then , now find $$[x^{10}](C(10,2)x^2 +C(10,3)x^3+C(10,4)x^4+C(10,5)x^5+C(10,6)x^6)^3$$
Understand the logic , when we expand it , we choose one term from each pharenteses where each pharenteses represent the groups. For example , one of the possible exponential selection is $3-2-5$ ,it means that we select $3$ people form the first group , $2$ people from the second group , $5$ people from the last group.So , one of the coefficient of $x^{10}$ is $C(10,3) C(10,2) C(10,5)$ .When we expand it with wolfram and look at the coefficient of $x^{10}$ ,we see the summation of all possible combinations.
*SEE EXPANSION *
So , answer is $24,466,050$
