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Let $g_n=$ number of 2-regular graphs on $n$ vertices

Let $c_n=$ permutations of $n$ with no fixed points or cycles of length 2

By a computation with the exponential generating function I think that the following formula should be true: $$c_n=\sum_{k=0}^n\binom{n}{k}g_kg_{n-k}\;.$$ There is a combinatorial proof of this fact?

The only thing that I noticed is that permutations that are cycles correspond to 2-regular connected graphs 2-to-1.

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up vote 3 down vote accepted

There seems to be a double-counting error in your result.

First off, it appears that by the number of $2$-regular graphs you mean what one would more precisely call the number of $2$-regular labeled graphs, whereas I believe the more usual interpretation of that phrase would be the number of $2$-regular unlabeled graphs, that is, the number of isomorphism classes of $2$-regular graphs.

The first interesting case is $n=6$. In this case, in addition to the $120$ $6$-cycles corresponding to $60$ cycle graphs, which your result gets right, there are some permutations and graphs that have two $3$-cycles. In both cases there are $\binom63$ ways to select the two groups of three that make up the cycles. In the case of the graphs, this completely determines the graph, so there are $\binom63$ such graphs. In the case of the permutations, there are two ways to orient the cycle for each cycle, for a total of four ways of orienting the two cycles, so there are $4\binom63$ such permutations. However, you result contains only one contribution in addition to the $2\binom63$ coming from $g_0g_6$ and $g_6g_0$, and that's $\binom63g_3g_3$, which is $\binom63$, so the result is one $\binom63$ short.

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The result is correct. You’ve double-counted: there are only $\binom52\cdot2^2$ permutations of $[6]$ that are products of two $3$-cycles, because there are $\binom52$ ways to fill out the $3$-cycle containing $1$. See this question and my answer. –  Brian M. Scott Oct 25 '13 at 9:20

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