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 ca+bnc \leq \frac{a+b}n . a,b,ca,b,c are all positive integers.

Find the number of ordered solutions of the 2n2n-tuplets of positive integers, (x1,x2,,xn,y1,y2,,yn)( x_1, x_2, \ldots, x_n, y_1, y_2, \ldots , y_n ) .

For example, if n=3,a=4,b=6,c=2n=3,a=4,b=6,c=2, then the number of ordered 6-tuplets of positive integers, (x1,x2,x3,y1,y2,y3)(x_1, x_2, x_3, y_1, y_2, y_3) is 168.

I wonder if it is possible to count mnm\cdot n tuples. For example, in this case, m=2m=2 because xi,yix_i, y_i have nn each, so 2 (=m)(=m) times nn is 2n2n tuples.

Note by Inquisitor Math
1 week, 5 days ago

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