Q:

How many rats are needed to find 3 poisoned bottles out of n bottles of wine? 

You organized a party with 1000 bottles of wine but you know that 1 bottle was poisoned before the party and you don’t want anybody to die. Luckily you are in the lab and you have 10 lab rats so you decide to use them to test which bottle is poisoned. The poison takes 1 hour to take effect also the party occurs in 1 hour. 

That was the original statement of the drunk rats problem. 
I was wondering how many rats you would need to need to detect for certain which 3 bottles of wine that are poisoned out of n bottles? 

A:

You can find one bottle from among $1000$ with $10$ rats because there are $1000$ possible one element subsets and $2^{10} > 1000$. So you number the bottles in base $2$ and the rats from $0$ to $9$ and give rat $r$ a sample from each bottles with a $1$ bit in place $r$.
For one bad bottle out of $n$ you need $\lceil \log_2(n)\rceil$ rats.  
To solve the $k$ bottle problem, number the  $N ={{n}\choose{k}}$ possible subsets of bad bottles, count them in binary.  You'll need $\lceil \log_2(N)\rceil$ bits, so that many rats.  
Caveat.  I'm pretty sure that will provide enough information to find the bottles, but I haven't thought through the proof in detail.  If I'm wrong I'm sure someone here will catch my error. 
Edit: Here's a reference from the OP's web page that points to a solution with fewer rats than mine.  So I still think I have enough information, but perhaps too much. 
https://mathoverflow.net/questions/59939/identifying-poisoned-wines
Edit (2): @Arby 's  comments below prompted this second edit.  I'm glad I was cautious making my naive claim.  It's easy to show it's  wrong.  With $2$ bad bottles in $4$ I predicted $3$ rats could find the bad pair.  If you write out my recipe for the $6$ possible pairs you'll find that all the rats die. 
Finally, I'm surprised that the OP accepted this wrong answer given that his question linked to a correct one.  At least I enjoyed solving the $1$ bottle puzzle, which I'd never seen.