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 $f(360, 20)$.

This is a tricky problem that involves deeper understanding in combinatorial mathematics.

Here we can regard it as arranging $\frac{m}{2}$ pairs of identical candles on a chandelier with $n$ candleholders where each pair is symmetrically arranged. Since the arrangement is symmetric, we regard the ring of candleholders as a circle and two arrangements that can be obtained from one another by a rotation as the same.

Thus, the problem is simplified to counting the number of ways to place $\frac{m}{2}$ indistinguishable pairs of objects into n indistinguishable boxes. This is a standard problem in combinatorics which can be solved by using generating functions, combinatorial identities or other combinatorial techniques.

The method using generating functions would be to associate a generating function to each box which is $1 + x^2 + x^4 + x^6 + …$ since a box can hold 0, 1, 2, … pairs of candles. And thus the generating function for $n$ boxes would be $(1 + x^2 +x^4 + x^6 + …)^n$, and the coefficient of $x^m$ in the expansion would give $f(n, m)$. But computing the coefficient in this way is very tedious and requires more advanced techniques for simplifying.

A better route would be using combinatorial identities to reach a closed expression. In this case, we would use the stars and bars theorem or the binomial coefficient theorem. The number of ways is given by $\binom{n}{m/2}$ when $n$ is even and $m$ is even and less than $n$ and by $0$ otherwise.

To summarize, to find $f(360, 20)$, we can use the formula described above:
$f(360,20)=\binom{360}{20/2}=\binom{360}{10}$

Then compute $\binom{360}{10}$, which is the number of ways to choose 10 items from 360. The computation itself might be a bit large and you might need to use a scientific or graphing calculator but that’s the principle of solution.

Thus, with comprehensive understanding in combinatorial mathematics, we can solve the problem. But you also need to pay attention to whether the numbers meet the conditions.

*Please note that this problem might require advanced knowledge beyond basic or intermediate math courses.

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