>**Proposition 1:** The number of integer solutions of the equation $$ \sum_{i=1}^{k}x_i = N $$ where $x_i\geq n_i$ for $i=1,\ldots,k$, is given by $$ \binom{N+k-1-n_1-n_2-...-n_k}{k-1} $$ Now, to tackle the problem as stated, you need to apply Proposition 1 and invoke the [inclusion-exclusion principle][1], in the following sense: For $i=1,...,k$, set as > $q_i$: the property of a solution of Proposition 1, to satisfy the condition $$ x_i> m_i $$ Then, if we denote: - $N(q_i)$, the number of solutions (provided by Prop. 1) satisfying property $q_i$, - $N(q_i q_j)$, the number of solutions (provided by Prop. 1) satisfying both properties $q_i$, $q_j$, .... etc, then we have (again applying Prop. 1) that: $$ N(q_1)=\binom{N-1-m_1-n_2-...-n_k}{k-1}, \\ N(q_2 q_3)=\binom{N-1-n_1-m_2-m_3-n_4-...-n_k}{k-1} $$ etc... Now all you need to do to obtain a compact formula for the number of solutions satisfying your constraints, is to apply the inclusion-exclusion principle to determine the number of solutions produced by Proposition 1, **which have none** of the properties $q_i$ for $i=1,2,...,k$. **Example:** As an example of application of the previous method, consider the following special case of the OP: >Find the number of (positive) integer solutions of the equation $$\sum_{i=1}^{k}x_i = N$$ for some positive integer $N$, given the constraints $1\leq x_i\leq \alpha$ for $i=1,\ldots,k$ The method described above gives: $$ {\small \binom{N-1}{k-1}+\binom{k}{1}\binom{N-\alpha-1}{k-1}+\binom{k}{2}\binom{N-2\alpha-1}{k-1}+\binom{k}{3}\binom{N-3\alpha-1}{k-1}+\cdots } $$ where $\binom{..}{..}$ stands for the [generalized binomial coefficients][2] and the summation halts when zero terms appear. [1]: https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle [2]: https://en.wikipedia.org/wiki/Binomial_coefficient