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In how many ways one can color n lined up bins in m colors with a constraint that colored bins are not allowed to be neighbors? Any number of bins can be left uncolored. any number of colors <= m can be used.

possibly a hint: I was going first to find all the combinations with one color and than allow each colored bin to have any color out of m for each combination.

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closed as off-topic by 3SAT, John B, Rory Daulton, hardmath, user26857 Feb 20 '16 at 21:25

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  • $\begingroup$ Are you required to use all $m$ colors? $\endgroup$ – Brian M. Scott Feb 20 '16 at 20:26
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    $\begingroup$ Welcome to MathSE! You are more likely to get a good answer to your question if you follow a few guidelines. In particular, what have you tried so far, and just where are you stuck? This is not a homework-answering site: we want to see that you have put significant work into the problem. $\endgroup$ – Rory Daulton Feb 20 '16 at 20:45
  • $\begingroup$ no, you can use any number of colors <= m $\endgroup$ – D.Luchinsky Feb 22 '16 at 1:02
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There might be a much easier way, but I see it this way:

Let $f(n, m)$ be the number of ways to color $n$ bins with $m$ colors and satisfy your condition.

$$f(n, m) = \overbrace{m \cdot f(n - 2, m)}^{\text{color the $n^{\text{th}}$ one, and skip one}} + \overbrace{f(n - 1, m)}^{\text{don't color the $n^{\text{th}}$ one}}$$

This order $2$ recurrence relation will require us to define $2$ base cases: $$f(1, m) = m + 1$$ $$f(2, m) = 2m + 1$$

Now since your $m$ is fixed, you can solve the following linear recurrence relation: $$f_n = m\cdot f_{n-2} + f_{n-1}$$

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Hint: You may consider that maximal number of bins to be colored is (n-1)/2 because if the number exceeds this two colored adjacent bins can occur

Hint2: For any subcase consider that the "moving" colored bin can have space relatively of how much do unclored=N/colored=C bins exist.

  • Example with colored C=(n-1)/2 there is no space for colored bins to move:

$NCNCNC...N$ so available moves for colored bins $C$ over non-colored $N$ are $0$ anywhere. $0000...$($(n-1)/2$ times)

$S_0=m^{\lfloor\frac{(n-1)}{2}\rfloor}$

  • Example with colored $C=(n-2)/2$ there is one bin space for each colored bin to move:

NCNNCNC...N ,or, NNCNCNNC...N the solution can be interpreted numerically as $010000...$ or $1000000...$ where $1$ means the bin is moved to right

NCNNCNC...N ,or, NCNCNNC...N this solution can be interpreted numerically as $010000...$ or $0010000...$

So on, ...we have one allowable move to either left or right for egding bins and 2 moves for the rest, in addition to the actual position, the formula for this case is written as $S_1=(1+1+2((n-2)/2-2)+1)*m^{(n-2)/2}$

...following this logic one can move on to $S_2$: $2000...$,$1100...$,$0200...$,$01010...$, ...etc

Hint3: From the numerical translation of the problem, we can extrapolate a general rule of $S_k$

$Sequence_k=x_1x_2....x_{(n-k-1)/2}$ where sum of $x$'s is $k$ and $x$ is between $0$ and $k$ inclusive.

The solution is the sum of all $S$'s

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