Problem 768

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 $f (360, 20)$ .

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 $f (360, 20)$ 。