Urn Permutations If one has 'plenty' of white, blue, and red balls and a collection of urns which can hold at most one ball. Then how many ways are there to arrange balls if
$1$. There are $n$ urns?
$2$. There are $n$ urns and no two adjacent urns can contain a red ball?
$3$. $n\geq 2$ and every urn with a blue ball must contain a red ball in one of the urns next to it.
The first is simple, it is $3^n$ possible combinations. It's the next two I have trouble with. The second sounds easier. If I have $n$ urns, I can go across and count the number of possibilities slowly. If I come to an urn with a red ball, there are only two choices for the urns next to it because neither can be red. Then I simply add up all these types of arrangements correct? 
The last one has me stumped. I thought I could count the cases where if there is a blue ball then there is a red ball to its left, then the right, then both and just eliminate the duplicates but this combined with the total $n$ possible slots turned out to be near impossible not to over/under count. Any advice on how to think about this counting problem to make it 'simpler'? 
EDIT: I mean there are $4^n$ total possible combinations because one could leave an urn empty.
 A: $2.$) We do not allow empty urns. That does not change the analysis:  just imagine that there is a fourth "colour" of ball,  transparent. 
Let $a_n$ be the number of arrangements with $n$ urns, and no two consecutive reds (good arrangements). 
How many good arrangements are there for $n+1$ urns? Let $n\ge 1$. A good arrangement for $n+1$ urns can be of two types: (i) the rightmost ball is non-red or (ii) the rightmost ball is red.
Type {i} arrangements can be obtained by appending any of the three non-red "colours" to a good arrangement of length $n$. So there are $3a_n$ of these.
Type (ii) arrangements can be obtained by appending a red to a good arrangement of length $n$ that does not end in red. That arrangement is obtained by adding a non-red to any good arrangement of length $n-1$. So there are $3a_{n-1}$ of these.
It follows that
$$a_{n+1}=3a_n+3a_{n-1}.$$
Solve this recurrence, with initial conditions $a_0=1$, $a_1=4$  in any of the usual ways.  For example, the characteristic polynomial of the recurrence is $x^2-3x-3$, with roots $\alpha=\frac{3+\sqrt{21}}{2}$ and $\beta=\frac{3-\sqrt{21}}{2}$. Then the solutions of the recurrence have shape $a_n=A\alpha^n +B\beta^n$, where $A$ and $B$ are constants. Find $A$ and $B$ so that the initial conditions are met. 
Remark:  An analysis along the lines of the answer to Question $2$ will yield a recurrence for Question $3$. 
A: Visualise this question in the form of slots and balls have to go in those slots.If we have an urn containing red ball the other two must not have red balls.That's correct,but we don't know about about other urns.They may or may not have red balls adjacent to each other.I find this problem quite easy if I use the inclusion -exclusion principle otherwise it's quite difficult.The following summation is your answer
3^n-(n-1)3^n-2+(n-2)3^n-3-(n-3)3^n-4+........+(-1)^n-1
Explanation
Total permutations are 3^n
But in these we have counted those cases when two red balls are adjacent so we
Subtracted them(the second term of the series).The case of arrangement of two adjacent red balls is nothing but arrangement of n-1 slots containing two red balls multiplied by other 3^n-2 cases as there are n-2 slots remaining.
But in the case of two adjacent red balls we have also subtracted those cases when 3 red balls are adjacent.So we added them.But again we have added the case of4 balls.So we had to subtract them.Please note that all cases of adjacent balls must be subtracted for only once.
