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In Blom, Holst, Sandell, "Problems and snapshots from the world of probability" there is a claim that the number of ways of placing $j$ dominos in a ring with $2n$ places in such a way, that each domino occupies two adjacent places, and the dominos do not overlap is $$ N_{2n,j} = \frac{2n}{2n -j} \binom{2n-j}{j} $$

How does one derive this result?

Sasha
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2 Answers2

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Corrected 30 September 2016.

Imagine that the $j$ dominoes have been placed. Paint one of them red, and break the ring immediately to its left. You now have a row of $2n$ places, $2j$ of which are covered by dominoes, including in particular the first two places. Number the dominoes from left to right, starting with the red one: $D_1,D_2,\dots,D_j$. The $n-2j$ uncovered places occur in $j$ blocks, some of which may be empty: after $D_j$, and between $D_k$ and $D_{k+1}$ for $k=1,\dots,j-1$. Counting the number of ways to distribute $2n-2j$ indistinguishable objects $-$ the uncovered places $-$ amongst $j$ distinguishable containers $-$ the spaces between adjacent dominoes and after $D_j$ $-$ is a straightforward stars-and-bars problem, and the answer is

$$\binom{(2n-2j)+j-1}{j-1}=\binom{2n-j-1}{j-1}\;.\tag{1}$$

However, the red domino could have occupied any of $2n$ different positions around the ring, so each of these arrangements actually corresponds to

$$2n\binom{2n-j-1}{j-1}\tag{2}$$

arrangements of one red and $j-1$ plain dominoes around the ring. On the other hand, in each arrangement of $j$ plain dominoes around the ring we could have chosen any of the $j$ dominoes to be the red one, so $(2)$ actually counts each arrangement of the plain dominoes $j$ times. The final result is therefore

$$N_{2n,j}=2n\cdot\frac1j\binom{2n-j-1}{j-1}=2n\cdot\frac1{2n-j}\binom{2n-j}j=\frac{2n}{2n-j}\binom{2n-j}j\;.$$

If you’re not familiar with the identity $$\frac1k\binom{n-1}{k-1}=\frac1n\binom{n}k\;,$$ expand both sides in terms of factorials. In the form $n\binom{n-1}{k-1}=k\binom{n}k$ it also has a very easy combinatorial proof: both sides calculate the number ways to choose a team of $k$ from a pool of $n$ players and appoint one member of the team to be its captain.

Brian M. Scott
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  • At (2), if we substitute $n=4, j=2$, it doesn't come out as an integer. How am I supposed to interpret this? – Junghyun Lee Sep 30 '16 at 05:45
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    @Nick: The correct interpretation is that I made a mistake; I’m amazed that no one caught it before now. Although the final result was right, the explanation of the intermediate steps was out of order, and the old intermediate expression $(2)$ to which you refer did not in fact represent what I said it did; as you observe, it isn’t even always an integer. I’m very glad that you caught drew my attention to it; I’ve now corrected the explanation (and fixed a moderately serious typo that everyone also seems to have missed). – Brian M. Scott Sep 30 '16 at 08:21
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Your problem is equivalent to selecting $j$ non-consecutive places on a circle of size $2n$. The formula $$N_{2n,j} = \frac{2n}{2n -j} \binom{2n-j}{j} $$ is explained here.

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