An officer has to solve a case with 20 suspects, 10 from colony A, and 10 from colony B. He can solve the case once at least 19 of them answer truthfully during an investigation.
The officer has two identical boxes labeled P and Q, which each have 20 cards, one for each suspect. Before each investigation, he takes one card from each box. He interrogates these two people during the investigation; the suspect from box P will tell the truth, and the suspect from box Q will tell the truth if and only if the suspect from box P is from colony A. (The officer can tell who's telling the truth.)
After each investigation, the officer will discard cards from truthful suspects (from both boxes) and return cards from lying suspects to the original box.
Find the number of possibilities that he can solve the case in 10 investigations.
It has been a week since this puzzle was first posted, and we seem uncertain of many facets of the scene presented. Since someone classified this as a "just math" problem, I suppose the attempt to give a combinatorial twist may be on the right track, though there are questions of "logic" (giving meaning to parts of the text) as well. Perhaps the phrase "return cards from lying suspects to the original box" conceals some clues we are missing. We have two "original boxes" , and we are not told specifically what "returning to the box" means -- is such a card placed on top of the stack, on the bottom, or where (if the first, there might be a sequence where the same suspect could be interrogated over and over).
As I stated in an earlier posting, it seems that as stated there is an obvious path so that the officer always gets the truth (i.e. the name(s) of the guilty person(s)) on the very first round of interrogation. I still see no obvious objection to that.