How many solutions are there to $x_1 + x_2 + ... + x_n = S$ for given $S$ assuming that $0 \le x_i \le 25$? I can solve it when there are no constraints on $x_i$ other that $x_i$ must be non-negative.
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what are the constraints on $n$ or what is its relationship with $S$? – Hypergeometricx Nov 05 '16 at 15:12
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This is a case for either "Inclusion-Exclusion" or "generating functions." Do you know either technique? – Thomas Andrews Nov 05 '16 at 15:15
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And yes, does $n$ vary, or are $S$ and $n$ fixed? – Thomas Andrews Nov 05 '16 at 15:16
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I've only heard about Inclusion-Exclusion principle, but don't know how this could be applied here. $n$ and $S$ may vary, I want to know how to solve general case, but you may assume that $S > n$ – J. Abraham Nov 05 '16 at 15:21
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When I say that $n$ is fixed, I mean, you don't want the sum of all values $x_1+x_2=S, x_1+x_2+x_3=S, \dots$, but that you want to just know the count for one pair $S,n$. – Thomas Andrews Nov 05 '16 at 15:27
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Yours is actually a special case of the earlier question (and of this one as well. You may also be interested in this question, in which the different variables can have different upper bounds. – Brian M. Scott Nov 05 '16 at 16:16
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This is a $\infty$-plicate. – Felix Marin Nov 06 '16 at 03:42
1 Answers
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Assuming $n$ fixed.
Generating function approach
Let $a_s$ be the number of solutions for $x_1+\cdots+x_n=s$ with $0\leq x_i\leq 25$. Then:
$$\begin{align}\sum_{s=0}^\infty a_sz^s &= (1+z+z^2+\cdots+z^{25})^n\\ &=\frac{(1-z^{26})^n}{(1-z)^n} =(1-z^{26})^n\sum_{i=0}^{\infty}\binom{i+n-1}{n-1}z^i \end{align}$$
Using $(1-z^{26})^n=\sum_{j=0}^n \binom{n}{j}(-1)^jz^{26j}$, you get:
$$a_S = \sum_{j=0}^{n} (-1)^j\binom{n}{j}\binom{S+n-1-26j}{n-1}$$
Inclusion-Exclusion approach
Let $A$ be the set of all solutions $x_1+\cdots+x_n=S$, with only the condition that the $x_i\geq 0$.
Let $A_i$ be the elements of $A$ with $x_i\geq 26$.
Then you want the size of $A\setminus(A_1\cup A_2\cup\cdots\cup A_n)$. This is ideal for inclusion/exclusion.
You get:
$$\begin{align}|A\setminus(A_1\cup\cdots\cup A_n)|=&|A|\\&-(|A_1|+|A_2|+\cdots+|A_n|)\\ &+(|A_1\cap A_2|+|A_1\cap A_3|+\cdots +|A_{n-1}\cap |A_n|)\\ &\vdots\\ &+(-1)^n|A_1\cap A_2\cap \cdots \cap A_n| \end{align}$$
This will turn into the same formula as above. Each term in each row is the same, and there are $\binom{n}{j}$ terms in the $j$th row (starting with $j=0$ at $|A|$.)
Thomas Andrews
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