Question: I have sixteen students in my class who sit in four rows of four. Each week they sit in a different order. 

After a number of weeks every student has sat next to every other student, next meaning side by side, one behind the other, or sitting diagonally together. What is the fewest number of weeks in which this can happen?

How many if my students were 25?

Answer: 
Partial answer to Question 1.

> 
> First, the average number of friends a student gets per week is
> $$\frac{4\times 3 +8\times 5+4\times 8}{16}=5.25$$
>  (there are $4$ corners each of which has $3$ friends, and so on).  
> 
>   
> 
> At the end of the day, we want each student to have had $15$ different friends, so this has to take **at least $3$ weeks.**
> Moreover, in three weeks at most $12$ students can be in the center once:
>  [![enter image description here](https://i.stack.imgur.com/RaUDE.jpg)](https://i.stack.imgur.com/RaUDE.jpg)  
> 
>   
> 
>  Those who are never in the center will need to be on an edge all three times to get $3\times 5$ friends.  
> 
>   
> 
>  The scheme above is therefore the only way to get 15 friends for everyone in 3 rounds. It is very rigid - there is no room for redundancy for the all-edgers, and there can be one redundancy for the others. Yet I can't prove yet that the constraints are too heavy to be met.
> 
> 
>
The first question i believe it's

> 
> 256
> 
> 
> 

 and the second one it's

> 
> 625
> 
> 
> 

I think the solution of this one is to multiply the two numbers given. An other example is:find all the combinations of these words `PANEL`(wich i'll use the same equation).