Problem Statement:-
In how many ways $10$ Indians, $5$ Americans and $4$ Englishmen can be seated in a row so that neither Americans nor Englishmen sit between Indians.
Attempt at a solution:-
$1^{\text{st}}$ Attempt:-
Since no Americans and Englishmen are to be seated between the Indians, so lets constraint the Indians to sit together and assume them to be a single entity. So, the total no of entities to be seated in a row become $5(\text{Americans})+4(\text{Englishmen})+1(\text{All Indians})=10$
So, total ways in which these entities can be permuted along a row= $P_{10}^{10}=10!$
But, since we had only assumed all the Indians to be one entity but since they are different, so they need to be arranged among themselves too.
So the total ways of arranging the Indians, Americans and Englishmen according to the problem statement $=10!\cdot 10!$
$2^{\text{nd}}$ Attempt:-
The firs attempt was the standard way of doing this. But I thought of doing this with the help of generating functions, as you will see below.
Consider the following arrangement of Indians and spaces between them:-
$$x_1\;\;\text I_1\;\;x_2\;\;\text I_2\;\;x_3\;\;\text I_3\;\;x_4\;\;\text I_4\;\;x_5\;\;\text I_5\;\;x_6\;\;\text I_6\;\;x_7\;\;\text I_7\;\;x_8\;\;\text I_8\;\;x_9\;\;\text I_9\;\;x_{10}\;\text I_{10}\;\;x_{11}$$
So, we get the following equation to solve
$$\sum_{i=1}^{11}{x_i}=9$$
Assuming Englishmen and Americans as identical for solving the equation.
We have the following conditions to take into consideration too $$0\le x_1,x_{11}\le 8\\ \qquad \qquad \qquad \qquad \qquad\; 0 \le x_r \le 9\;,\qquad r=\{2,3,4,5,\ldots ,10\}$$
So, we have the following generating function in which we have to find the coefficient of $x^9$
$$\begin{aligned} &\text{coefficient of $x^9$}(1+x+x^2+\ldots+x^8)^2(1+x+x^2+x^3+\ldots+x^9)^9\\ =&\text{coefficient of $x^9$}\left(\frac{1-x^9}{1-x}\right)^2\left(\frac{1-x^{10}}{1-x}\right)^9\\ =&\text{coefficient of $x^9$}(1-x)^{-11}\\ =&\binom{11+9-1}{9}=\binom{19}{9} \end{aligned}$$
Now, since the Englishmen and Americans are distinct despite assuming them to be indistinguishable and the Indians can be arranged among themselves, we get the total ways to be
$$\binom{19}{9}\cdot 10!\cdot 9!=19!\text{(which is the total ways of arranging without any restriction)}$$
My deal with the question:-
In my $2^{\text{nd}}$ attempt I think that the way that I assumed the range of values of $x_1, x_{11}$ is creating the problem. If that is not so, I don't know what is it that is going wrong here, and if it so happens that the range of values of $x_1,x_{11}$ is creating the problem then also I am not able to come up with the right thing to do.
Also, I know that an alternative approach for this method (using generating functions) would have been assigning the range as follows
$$0\le x_1,x_{11}\le 9\\ \qquad \qquad \qquad \qquad\qquad \qquad \qquad x_r=0 \;,\qquad \qquad \qquad r=\{2,3,4,5,\ldots ,10\}$$
But that would have been same as my $1^{\text{st}}$ Attempt