Combinatorics - multi-group question - Choose 6 people out of 14, respecting conditions. Here's the question:
"A congress will be attended by two representatives from Colombia, three from Chile, four from Argentina and five from Brazil.  Each of the $14$ representatives prepared their own speech, but only $6$ will be drawn to speak.  If the draw rule provides that each of the four countries must have at least one representative speaking, the number of different ways to compose the set of six speeches that will be heard at the congress, regardless of the order, is equal to how much?"
I've been working on this question for two weeks. My answer is $1450$ - but I solved it via brute force. I literally counted all the possibilities of the rule being broken and subtracted from the $3003$ possibilities to form groups of 6 people out of $14$.
I tried to look at the problem with fewer countries, and fewer choices. It helped me to count, but I couldn't decipher the pattern with which the problem changes when adding a new representative or a new country.
Can someone help me?
 A: A nice way to do this is to use the principle of inclusion-exclusion. Take all $\binom{14}6$ ways to choose $6$ people, then for each country, subtract the assignments where that country has no chosen representatives. The result is
$$
\binom {14}6
-\underbrace{\binom{12}6}_{\text{Columbia missing}}
-\underbrace{\binom{11}6}_{\text{Chile missing}}
-\underbrace{\binom{10}6}_{\text{Argentina missing}}
-\underbrace{\binom{9}6}_{\text{Brazil missing}}
$$
However, there is a problem. Arrangements with two missing countries will be subtracted twice in the above computation. To fix this, we must add these doubly-subtracted arrangements back in. To the above, we add
$$
+\underbrace{\binom{9}{6}}_{\text{Columbia + Chile}}
+\underbrace{\binom{8}{6}}_{\text{Columbia + Argentina}}
+\underbrace{\binom{7}{6}}_{\text{Columbia + Brazil}}
+\underbrace{\binom{7}{6}}_{\text{Chile + Argentina}}
+\underbrace{\binom{6}{6}}_{\text{Chile + Brazil}}
$$
At this point, we are done! We would not be done if there were any arrangements counted by $\binom{14}6$ with three countries missing, but that is not possible.
A: Fo a second way , use generating functions. If there is at least one president in each country , then
Genereting function of Columbia : $$\binom{2}{1}x^1 +\binom{2}{1}x^2$$
Genereting function of Chile :
$$\binom{3}{1}x^1 + \binom{3}{2}x^2 +\binom{3}{3}x^3$$
Genereting function of Argentina :
$$\binom{4}{1}x^1 + \binom{4}{2}x^2 +\binom{4}{3}x^3 + \binom{4}{4}x^4$$
Genereting function of Brasil :
$$\binom{5}{1}x^1 + \binom{5}{2}x^2 +\binom{5}{3}x^3 + \binom{5}{4}x^4 +\binom{5}{5}x^5$$
Now, find the expansion of them and find the coefficient of $x^6$
Calculation in this link
So , answer is $1450$ , you are right !.By the way the binomial coefficients means the number of selection among candidates.For example , $\binom{5}{2}x^2$ means number of selection of two president among $5$ in Brasil.
Do you get the idea ?
$\mathbf{\text{MORE CLARIFICATION:}}$ In given link ,you will see that binomial expansions like $((1+x)^2 -1)$ ,  we know that when we expand it , it will give $2x +x^2$ . I wanted to write them clear form , but wolfram could not calculate them because of the lenght of expansion.Secondly , by the form of generating function (we started from $x^1$) , the number of president more than $3$ is automatically prevented.
A: $$\sum_{\sum_{i=1}^4 \alpha_i=6\\1\leq \alpha_i \leq i+1}\binom{2}{\alpha_1}\binom{3}{\alpha_2}\binom{4}{\alpha_3}\binom{5}{\alpha_4}=1450$$
