UNSW MathSoc Championship Q1

Using the Inclusion-Exclusion Principle and a bit of tinkering, this problem can be generalised. Namely, the number of ways $N$ of arranging $n$ distinctly coloured pairs of identically coloured marbles in a line such that at least one pair of adjacent marbles is an identically coloured pair can be given by $N=\sum_{k=1}^n(-1)^{k-1}\binom{n}{k}\dfrac{(2n-k)!}{2^{n-k}}$ Substituting $n=2$ yields $N=4$ , making the answer $\boxed{4}$ .