In how many ways are we able to arrange $k$ identical non-overlapping dominoes on a circle of $2n$ labelled vertices?

Question

In how many ways are we able to arrange $k$ identical non-overlapping dominoes on a circle of $2n$ labelled vertices?

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The problem can be reduced to the number of ways to choose $k$ non-consecutive vertices from the $2n$ labelled vertices,and now there are two cases,assuming we are arranging these $k$ identical dominoes counterclockwise:

• If vertex $1$ in included,then the adjacent vertex (from the left) cannot be chosen,since the dominoes are non-overlopping,so we are left with the other $2n-2$ vertices and we want to choose $k-1$ nonconsecutive vertices ,this can be done in $\binom{2n-2-(k-1)+1}{k-1}=\binom{2n-k}{k-1}$ ways.

• If vertex $1$ in not included,so we are left with the other $2n-1$ vertices and we want to choose $k$ nonconsecutive vertices ,this can be done in $\binom{2n-1-k+1}{k}=\binom{2n-k}{k}$ ways.

Now summing these two cases gives the answer.

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I'm not sure about the proof,besides dos it make difference if we do the process clockwise?

Answer 1

You appear to be off a bit: in your first case $3$ vertices are unavailable, not $2$.

I’ve numbered the vertices from $1$ through $2n$. For my first case I put a domino on vertices $1$ and $2$. Now I need to choose $k-1$ of the $2n-3$ vertices $3,4,\ldots,2n-1$, ensuring that no two chosen vertices are adjacent. This can be done in

$$\binom{(2n-3)-(k-2)}{k-1}=\binom{2n-1-k}{k-1}$$

ways.

For my second case I put a domino on vertices $2n$ and $1$; the analysis is the same, so we get another $\binom{2n-1-k}{k-1}$ arrangements.

Any other arrangement must avoid vertex $1$ entirely. In that case we need to choose $k$ of the $2n-2$ vertices $2,3,\ldots,2n-1$, ensuring that no two chosen vertices are adjacent. This can be done in

$$\binom{(2n-2)-(k-1)}k=\binom{2n-1-k}k$$

ways. I get a total of

$$2\binom{2n-1-k}{k-1}+\binom{2n-1-k}k=\binom{2n-1-k}{k-1}+\binom{2n-k}k$$

arrangements. I’ve checked this by hand with $n=4$ and $k=3$.

Answer 2

Let's use $d_k$ to denote the number non-overlapping arrangements of $k$ dominoes on a circle of $2n$ vertices.

Your calculation contains an error. If vertex $1$ is included, then three vertices must be excluded, not two. Since the domino will cover vertices $1$ and $2$, neither of these may be included. In addition, as you point out, vertex $2n$ must be excluded. So the correct result is $$ d_k=\binom{2n-k-1}{k-1}+\binom{2n-k}{k}. $$

Here's an alternative method. Let's place a token on each vertex that is not covered by a domino. So every vertex is now covered either by a token or by a domino and the number of dominoes and tokens taken together is $k+2(n-k)=2n-k$.

Let's now make a special mark on one object, either domino or token. Since there are $2n-k$ objects we could choose to mark, the number of marked arrangements is $(2n-k)d_k$.

Marked arrangements can be counted in a different way, however, and equating the two answers will allow us to find $d_k$. This second method is to lay the $k$ dominoes and $2(n-k)$ tokens out in a line rather than a circle. There are $\binom{2n-k}{k}$ ways to do this. The first object in the line will be the marked one. Now we wrap the arrangement on the circle. Let the vertices be numbered $1$ to $2n$. We just need to pick one of these vertices to be the starting point. So there are $2n\binom{2n-k}{k}$ marked arrangements.

Equating this to our previous answer shows that $$ d_k=\frac{2n}{2n-k}\binom{2n-k}{k}. $$ Since $\binom{2n-k-1}{k-1}=\frac{k}{2n-k}\binom{2n-k}{k}$, the two methods agree.