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Problem 768

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Chandelier

A certain type of chandelier contains a circular ring of $n$ evenly spaced candleholders.

If only one candle is fitted, then the chandelier will be imbalanced. However, if a second identical candle is placed in the opposite candleholder (assuming $n$ is even) then perfect balance will be achieved and the chandelier will hang level.

Let $f(n,m)$ be the number of ways of arranging $m$ identical candles in distinct sockets of a chandelier with $n$ candleholders such that the chandelier is perfectly balanced.

For example, $f(4, 2) = 2$: assuming the chandelier’s four candleholders are aligned with the compass points, the two valid arrangements are “North & South” and “East & West”. Note that these are considered to be different arrangements even though they are related by rotation.

You are given that $f(12,4) = 15$ and $f(36, 6) = 876$.

Find $f(360, 20)$.

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枝形吊灯

某种枝形吊灯上等距离安置了一圈共$n$个烛台。

如果只摆上一根蜡烛，那么吊灯将会失去平衡。必须在与第一根蜡烛相对的烛台上（假设$n$是偶数）摆上第二根完全相同的蜡烛，才能让吊灯保持平衡。

对于安置有$n$个烛台的吊灯，记$f(n,m)$为在不同的烛台上摆上共$m$根完全相同的蜡烛且仍能让吊灯保持平衡的方案总数。

例如，$f(4, 2) = 2$：不妨假设吊灯上的四个烛台恰好指向正东、正西、正南、正北方向，那么符合要求的两种方案是“正北和正南”以及“正东和正西”。注意尽管这两种方案可以经过旋转重合，但仍视为不同的方案。

此外，还已知$f(12,4) = 15$以及$f(36, 6) = 876$。

求$f(360, 20)$。

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