Hi,
I think
LCKurtz talks about Counting Principle as in the book. Yes its possible: For Doors he has : 10 * 9 options or days i.e after 90 days there is a possibility that he is going to perform a repeat enterance from the same and leave from the same i.e. the duo is same as he did once in the last 90 days. Let he has has Doors D1..D10.
So first day he entered through D1 & leave through D2. Second day he entered again through D1 but left through D3. Thus for 1st nine days he kept entering from D1 but choose other 9 doors for leaving each time he left with a unique door. On 10th day he entered through D2 but choose D1 & D3..D10 for leaving, keeping in view the uniqueness factor for the next 9 days. Thus on 91st day there is a possibility of repeatation. That is he enters from D1 & leaves through D2. So we won't consider the remaining options.For Stair cases, in the same way, on the first he can chose from any of the 1-8 staircases(S1-S8) to climb up & go down through1-7 staircases in order to avoid repeatation. Let suppose he went up through S1 & came down through S2. On the second day he went up through S1 & came down through S3. So he has following combinations without repeatation for first 7 days: {S1-S2, S1-S3, S1-S4, S1-S5,S1-S6, S1-S7, S1-S8} On the 8th day he can choose S2 fopr up & S3 for down so he has following climbing pattern for next seven days:{S2-S1, S2-S3, S2-S4, S2-S5, S2-S6,S2-S7, S2-S8}. Thus for first 56 days no repeation occurs but 57 day, there is a repeatation.
If we limit by staircases the he can have only 56 days without repeatation but if we consider the whole path from Door In- Stair Up-Stair Down- Door out then he can have:
10 * 9 * 8 * 7 days with repeatation of any path.This is the answer in the book also. Thanks everybody.Zulfi.