# 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 $c \leq \frac{a+b}n$. $c$ is a positive integer.

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
3 months, 2 weeks ago

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There is no bound for C, that means infinite solutions for the ordered set. It says $x_i+y_i≥c$, multiply the whole equation by $n$ and you get $nx_i+ny_i≥c$ which would mean $a+b≥nc$, divide the whole equation by $n$ and you have $c≤ \dfrac{a+b}{n}$, c could be any number and you can list such positive integers, greater than c .

I don’t quite get what you want to say. a,b,c are some integers, not all integers. Like the example above, there is a certain number of solutions for some a,b,c.

By the way, thx for coming by :)

- 3 months ago

If my doubt is gibberish you can for sure tell me, or maybe you really did not understand what I'm trying to say. Just do share when you find the right solution, or if you already have it.

I do not have a solution for this. That is why I want to ask this problem.

I would like to know what you mean by 'you can list such positive integers, greater than c .'

By the way, have you tried out the example? That might give some sense what this problem is about

- 3 months ago