What is the number of ways to put 40 identical balls in 6 boxes while no box contains more than 14 balls. I believe that I have to use the Inclusion–exclusion principle but I am not sure how. I tried to use this formula: $\binom{n+k-1}{n-1}$ for every step of the inclusion–exclusion solution but it didn't work out. 
Thank you!!

Jack D'Aurizio gives the clue, it can be solved by using the generating functions. In this case, generating function is
$$G(x) = (1+x+x^2+\ldots+x^{14})^6$$
and we are seeking the coefficient of $x^{40}$. Now, if we manipulate the expression by using:
$$1+x+x^2+\ldots+x^{14} = \frac{1-x^{15}}{1-x}\ and\ \frac{1}{1-x} = 1+x+x^2+...$$
we get
$$G(x) = (1-x^{15})^6 \cdot (1+x+x^2+...)^6$$
Now, we also know that we have
$$(1+x+x^2+...)^6 = \sum_{m = 0}^{\infty}\binom{6+m-1}{6-1}x^m$$
So generating function becomes
$$G(x) = (1-x^{15})^6 \cdot \sum_{m = 0}^{\infty}\binom{5+m}{5}x^m$$
So the coefficient of the term $x^{40}$ is:
$$\binom{6}{0} \binom{45}{5}-\binom{6}{1}\binom{30}{5}+\binom{6}{2}\binom{15}{5}$$
