Question:We have a box with 50 number balls and we take out 6 of them (without putting any of them back). If $X=\textit{minimum}$ and $Y=\textit{maximum}$ then what is $f_{XY}(x,y)=P(X=x \cap Y=y)$ ?
Attempted Sollution: Obviously if $y-x\leq4$ then the $f_{XY}(x,y)=0$. For the non trivial part,where $y-x\geq 5$: $f_{XY}(x,y)=\frac{{y-x-1}\choose 4}{50\choose 6}$. This is because all the possible outcomes are $50\choose 6$ (since we don't care about the order) and the for desired outcomes we want to pick 4 out of all the balls between $x,y$ which are $y-x-1$.
Now, in my notes, the suggested answer is $5\cdot6\cdot\frac{{y-x-1}\choose 4}{50\choose 6}$. Which answer is correct and if mine is wrong, where is the fault ?