# Four colors, no back-to-backs (recursive)

#### jflurrie

(I posted already that for-loop question from our exam, sorry if this comes out as flooding)

So the question was short and clear, but I can't figure out the answer.

We have four colors (blue and three others) of poker chips in a stack of n chips. In how many ways can the stack be formed so that blue chips are not directly on top of each other?

What I understood by myself:
For example with three chips the options are
1) b o b (o = other, b = blue)
2) o b o
3) o o b
4) b o o
5) o o o

Which gives us 3+9+9+9+27=57 ways.

With two chips the options are
1) b o
2) o b
3) o o

Which gives us 3+3+9 = 15 ways.

But I can't figure out any kind of formula for this.

#### Plato

MHF Helper
I cannot give you a recursive solution. But can give a counting solution.
If we have $$\displaystyle N$$ chips then maximum number of blue chips is $$\displaystyle \max = \left\lceil {\frac{N}{2}} \right\rceil$$.
Then the count is $$\displaystyle \sum\limits_{k = 0}^{\max } {\binom{N-k+1}{k}3^{N - k} }$$.

#### jflurrie

I cannot give you a recursive solution. But can give a counting solution.
If we have $$\displaystyle N$$ chips then maximum number of blue chips is $$\displaystyle \max = \left\lceil {\frac{N}{2}} \right\rceil$$.
Then the count is $$\displaystyle \sum\limits_{k = 0}^{\max } {\binom{N-k+1}{k}3^{N - k} }$$.
The max part is clear - if you have more than half of stack blue chips, there has to be blue chips on top of each other. On count, that 3 is the number of other colors, right?

#### Plato

MHF Helper
The max part is clear - if you have more than half of stack blue chips, there has to be blue chips on top of each other. On count, that 3 is the number of other colors, right?

#### awkward

MHF Hall of Honor
Hi jlflurrie,

Plato has given you an excellent solution, but if I understand you correctly, you may be interested in a recursive solution, is that right?

If so let's say an arrangement is "acceptable" if no blue chips are adjacent, and let $$\displaystyle a_n$$ be the number of acceptable arrangements with n poker chips. We know

 $$\displaystyle a_1 = 4$$ and
 $$\displaystyle a_2 = 15$$.

Now suppose we have a stack of n+1 chips, where $$\displaystyle n \geq 2$$. The chip on top is either blue or one of the other colors. If it is blue, then the chip under it must be one of the other 3 colors, and the remaining n-1 chips can be any acceptable arrangement; so the number of arrangements with a blue chip on top is $$\displaystyle 3 a_{n-1}$$. If the chip on top is one of the other 3 colors, then the remaining n chips can be any acceptable arrangement; so this can be done in $$\displaystyle 3 a_n$$ ways.

Hence
$$\displaystyle a_{n+1} = 3 a_n + 3 a_{n-1}$$.

Together with  and , this is enough to find $$\displaystyle a_n$$ for any n.

Last edited:
• Plato