7. The number of unordered solutions $\left[n_{1}, \ldots, n_{N}\right]$ of the system $n_{1}+\ldots+n_{N}=n, n \geqslant 1, N \geqslant 1$, with non-negative integers $n_{i} \geqslant 0$, $i=1, \ldots, N$. What is the number of such solutions? What is the number of such solutions if only positive solutions $\left(n_{i}>0\right.$, $i=1, \ldots, N$ ) are allowed? What is the number of ordered solutions ( $n_{1}, \ldots, n_{N}$ ) of the same system $n_{1}+\ldots+n_{N}=n, n \geqslant 1, N \geqslant 1$, with positive $n_{i}$, $i=1, \ldots, N$ ?
Solution. Let the number of unordered non-negative solutions of the equation $n_{1}+\ldots+n_{N}=n$ be denoted by $R_{n}^{N}$. Then all solutions can be divided into those containing at least one zero - the number of such solutions is $R_{n}^{N-1}$ - and those that do not contain zeros - the number of such solutions is $R_{n-N}^{N}$ (since we can subtract 1 from each term and obtain a solution of the system with $n^{\prime}=n-N$). Thus, $R_{n}^{N}=R_{n}^{N-1}+R_{n-N}^{N}$, which, together with the boundary conditions $R_{n}^{1}=1$ for $n \geqslant 0$ and $R_{n}^{N}=0$ for $n<0$, allows us to quickly find $R_{n}^{N}$. However, an explicit formula for $R_{n}^{N}$, as far as we know, is unknown.

The number of unordered positive solutions of the same equation is $R_{n-N}^{N}$ - these are precisely the non-negative solutions that do not contain 0.

The number of ordered positive solutions of the given equation is exactly the same as the number of non-negative solutions of the equation $n_{1}+\ldots+n_{N}=n-N$, it is sufficient to subtract 1 from each number in each positive solution. Thus, in this case, the answer is $C_{n-1}^{n-N}$ by virtue of problem I.1.6(b).