I need help to solve this problem
an urn contains the numbers 1,2,3, ......., 2019. we draw at random, without replacement, four numbers in the order, in the urn, which we will designate by a, b, c , d. what is the probability that the system (S)
(S): ax+by=ab and cx+dy=cd
has a strictly internal solution (that is, not on the axes) in the first quadrant?
It can easily be found that $(x,y)=(\frac{bd(c-a)}{bc-ad},\frac{ac(b-d)}{bc-ad})$.
As it has a 'strictly internal solution' $c-a$, $b-d$ and $bc-ad$ must have the same sign.
If $c>a$, then $b>d$. Under this condition, $bc-ad$ is always true. Similarly, if $c<a$ and $b<d$, then $bc-ad<0$ is always true.
Now, the total number of possible relative positions of a,b,c and d under the given condition is $2\cdot\frac{4!}{(2!)^{2}}$ (Under each condition, the relative position of a,c and b,d are fixed). So, the total probability that the solution is strictly internal to the first quadrant is $\frac{2\cdot\frac{4!}{(2!)^{2}}}{4!}=\frac{1}{2}$.