Am I supposed to use generating functions or combinatorics or something for this question? If there are 201 seats in the Parliamentary chamber, how many different ways are there for the numbers
of seats to be allocated amongst three political parties, subject to no one party having an overall
majority?
I could figure it out manually. 201/3 = 67 per party. Thats one way. Then divide 67/2 (because you can move 2 from one party at a time to ensure the other 2 are equal) = 33.5 but use lower bound as a person cannot be split in half. That makes it 100+100+1 = 201. So now thats 33 ways. But party that gives up 2 each time can change. i.e. Party A gives up 2 members to B&C until 1 remain. Then party B gives up... then C. So would it be 1+(33*3) = 100 ways? 
Is there a cleaner way to do this, say using generating funtions or combinatorics of some sort? If not I dont know why the would ask this question.
Any help appreciated.  

We seek the number of integer solutions to the equation
$$
x_1+x_2+x_3=201\tag{1}
$$
where $ 0\leq x _i \leq 100$. You can solve this using generating functions by noting that this is equivalent to finding the coefficient of $x^{201}$ in the generating function of 
$$
(1+x+\dotsb x^{100})^3
=\frac{(1-x^{101})^3}{(1-x)^3}
=(1-3x^{101}+3x^{202}+x^{303})\left(\frac{1}{(1-x)^3}\right).\tag{2}
$$
Use the identity 
$$
\frac{1}{(1-x)^3}=\sum_{n=0}^\infty \binom{n+2}{2}x^n.\tag{3}
$$
(which can be obtained by repeatedly differentiating the geometric series) to find that the coefficient of $x^{201}$ is 
$$
\binom{201+2}{2}-3\binom{(201-101)+2}{2}\tag{4}
$$
which can be obtained equivalently using inclusion exclusion.
