# How many mn tuples are there??

Let $$\{x_1, x_2, \ldots , x_n \}$$ and $$\{y_1, y_2, \ldots, y_n\}$$ be sets of non-negative integers such that $$\sum \limits_{i=1}^n x_i = a$$, $$\sum \limits_{i=1}^n y_i = b$$, and $$x_i + y_i \geq c$$ for all $$i = 1,2,\ldots, n$$.

We are also given that $c \leq \frac{a+b}n$. $a,b,c$ are all positive integers.

Find the number of ordered solutions of the $2n$-tuplets of positive integers, $( x_1, x_2, \ldots, x_n, y_1, y_2, \ldots , y_n )$.

For example, if $n=3,a=4,b=6,c=2$, then the number of ordered 6-tuplets of positive integers, $(x_1, x_2, x_3, y_1, y_2, y_3)$ is 168.

I wonder if it is possible to count $m\cdot n$ tuples. For example, in this case, $m=2$ because $x_i, y_i$ have $n$ each, so 2 $(=m)$ times $n$ is $2n$ tuples. Note by Inquisitor Math
1 week, 5 days ago

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