Hi,
I have got the following question:

Josh works on the second floor of a building. There are 10 doors to the building and 8 staircases from 1st to the 2nd floor. Josh decided that each day he would enter by one door and leave by a different one and go up one stair case and down another. How many days could Josh do this before he had to repeat a path he had previously taken?

There are more doors and less stair cases. So we would stop when we would have a repeatation of Staircases. Answer should be in factorial. 10! for Doors which is the number of days but before these number of days we would stop because there are less number of stair case . So we have 10!/8!. Or it would be 8!/2! because we have 2 staircases per day.

Somebody please guide me.

Zulfi.

 

Hi,
I have got the following question:

Josh works on the second floor of a building. There are 10 doors to the building and 8 staircases from 1st to the 2nd floor. Josh decided that each day he would enter by one door and leave by a different one and go up one stair case and down another. How many days could Josh do this before he had to repeat a path he had previously taken?

There are more doors and less stair cases. So we would stop when we would have a repeatation of Staircases. Answer should be in factorial. 10! for Doors which is the number of days but before these number of days we would stop because there are less number of stair case . So we have 10!/8!. Or it would be 8!/2! because we have 2 staircases per day.

Somebody please guide me.

Zulfi.

Hi,

Why dont you try to think it out a little more. You can organize your thoughts as to what he might actually be doing each day.
For example, when you start figuring with the doors alone, if he goes in the same door every day for 9 days that means he goes in door 1 and exits on any door except door 1 for 9 days. Then if he switches to door 2 , he can go in that door for 9 days as he goes out the other 9 doors on separate days. Add up all the possibilities.
You can then do the stairs.

See if that helps.

 

Hi,
I have got the following question:

Josh works on the second floor of a building. There are 10 doors to the building and 8 staircases from 1st to the 2nd floor. Josh decided that each day he would enter by one door and leave by a different one and go up one stair case and down another. How many days could Josh do this before he had to repeat a path he had previously taken?

There are more doors and less stair cases. So we would stop when we would have a repeatation of Staircases. Answer should be in factorial. 10! for Doors which is the number of days but before these number of days we would stop because there are less number of stair case . So we have 10!/8!. Or it would be 8!/2! because we have 2 staircases per day.

Somebody please guide me.

Zulfi.

Why would you stop when you have a repetition of staircases? If I come in a different door but then use the same staircases and exit door, am I not using a different path than before.

Reduce the problem to three doors and two staircases and then list all of the possible paths. How many are there? Can you see a pattern that would let you break the problem into two pieces, one of which focuses on the doors and the other on the staircases?

 

Hi,

I think he 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 wont 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 with out 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.

 

Good.

Now see if you can tighten up your description and make it more general (i.e., you have D doors and S staircases).