Odds of getting y objects in one group when choosing x from a group of N objects without replacement

Odds of getting y objects in one group when choosing x from a group of N
objects without replacement

So the problem I'm trying to work out is: if you have a group of 30
objects and split it in half, what are the odds that 6 of the objects end
up in one group, except I also wanted the general solution.
I think I have it, but I'm not certain and wanted to check here, so here's
my thinking: Call the total number of objects $N$, the number that you're
selecting $x$, and the number of objects you're interested in $y$, and
then the total number of ways of getting $x$ objects from $N$ is $C(N,x)$.
To get the number of ways in which you have the $y$ objects in that group,
I essentially just say that there's a new group of $N-y$ objects (since
you've already chosen the other $y$ objects, but this is the bit of
thinking that I feel might be wrong) of which you have to choose $x-y$
objects to make up the rest of your selection of $x$. So then if that's
all correct, the odds of this situation happening is
$\frac{C(N-y,x-y)}{C(N,x)}$, which I wrote out and cancelled a factor of
$(N-x)!$ to get $\frac{x!(N-y)!}{N!(x-y)!}$, and this gives a probability
of about $8.4\times10^{-3}$ or 1 in 119 for the numbers above, which I
guess seems a little bit high but I'm not sure.