Only Integers!

The equation $3y-5x=246-xy$ is a Diophantine equation.

It can be rewritten as $3y-5x+xy-15=246-15$

Now, this equation can be nicely factorised,

$(y-5)(x+3)=231$

The various ways 231 can be written as the product of 2 positive integers are-

$231= 21*11, 77*3, 33*7, 231*1$ ( Note that 231= $3*7*11$ )

For each of those 4 unique ways that 231 can be factored, all of them will produce 2 solutions for (x,y) except $231*1$ which will produce only 1 solution (since (x+3) cannot be equal to 1).

Therefore the total solutions (x,y) possible are $(2*3)+1 = \boxed{7}$

$3y - 5x = 246 - xy$

$\rightarrow xy + 3y = 5x + 246$

$\rightarrow y(x + 3) = 5x + 15 + 231$

$\rightarrow y = \frac{5x +15}{x + 3} + \frac{231}{x + 3}$

$\rightarrow y = 5 + \frac{231}{x + 3}$

Therefore, $y$ is an integer if $x + 3$ is a factor of $231$ . $231$ has $8$ factors: $1$ , $3$ , $7$ , $11$ , $21$ , $33$ , $77$ , and $231$ . All of them give non-negative integers for $x$ and $y$ except for $1$ (when the factor is $1$ , $x = -2$ ), so there are $\boxed{7}$ non-negative integer solutions.