Number of ways to distribute 100 identical chairs among 4 different rooms In how many ways can 100 identical chairs be divided among 4 different rooms so that each room will have 10,20,30,40 or 50 chairs?
I'm having problems coming up with the generating function for this question.  
The answer given is 68. 

The problem can also be done without generating functions.
After dividing by $10$ we need to find the number of ordered quadruples of numbers in $\{1,2,3,4,5\}$ that add to $10.$  This is equivalent to the number of ordered quadruples of numbers in $\{0,1,2,3,4\}$ that add to $6.$  If we drop the restriction that the numbers be no greater than $4,$ we can use stars-and-bars to find that the number of quadruples is $\binom{6+3}{3}=84.$  
Now we need to subtract the number of quadruples that contain a number greater than or equal to $5.$  Such a quadruple contains exactly one such number, given that the numbers add to $6.$   We represent a quadruple in which the first number is greater than or equal to $5$ by $(5,0,0,0)+q,$ where $q$ is an ordered quadruple of non-negative integers that add to $1.$  We represent quadruples where the number greater than or equal to $5$ is in positions $2,$ $3,$ or $4$ similarly.  Since there are four choices of position, and four choices of $q,$ there are $16$ quadruples containing a number greater than or equal to $5.$  This gives the desired result.
Added: I should add that the calculation above can be seen in the generating function approach.  Using the expression in Thijs' answer, we get
$$
\begin{aligned}
\left[x^{10}\right]\left(x+x^2+x^3+x^4+x^5\right)^4&=\left[x^6\right]\left(1+x+x^2+x^3+x^4\right)^4\\
&=\left[x^6\right]\left(\frac{1-x^5}{1-x}\right)^4\\
&=\left[x^6\right]\frac{1-4x^5}{(1-x)^4}\\
&=\left[x^6\right]\left(1-4x^5\right)\sum_{j=0}^\infty\binom{-4}{j}(-1)^jx^j\\
&=1\cdot\binom{-4}{6}(-1)^6-4\cdot\binom{-4}{1}(-1)^1\\
&=\binom{9}{6}-4\binom{4}{1}.
\end{aligned}
$$
The notation $[\text{blah}]$ means "coefficient of blah".  In the third line we discarded numerator terms of order $x^{10}$ or higher.  In line four, we used the generalized binomial theorem to expand the denominator.  In line six, we rewrote the binomial coefficients $\binom{n}{r}$ with negative $n$ in terms of positive $n.$
If one traces through the steps of this calculation, one sees that they correspond with the steps of the previous calculation.
