Identify the number of distinct string of length 2 of the combinations of blue and yellow beads. The both ends of the strings are not marked due to this they are indistinguishable if interchanging the ends of one will results other. Denote \(b\) and \(y\) as a blue and yellow beads. Let us consider \(bb\), \(by\), \(yb\) and \(yy\) are the four different strings of length 2 when equivalence between strings is not taken into the consideration. The number of equivalent classes into which the set \(s = (bb, by, yb, yy)\) is divided by equivalent relation induced by the group of permutation \([\{\Pi_{1}, \Pi_{2}\}, 0]\) where : \(\Pi_{1} = \begin{pmatrix} bb & by & yb & yy \\ bb & by & yb & yy \end{pmatrix}\) \(\Pi_{2} = \begin{pmatrix} bb & by & yb & yy \\ bb & yb & by & yy \end{pmatrix}\)
3
The permutation \(\Pi_{1}\) is the identity, which leaves every string unchanged, and \(\Pi_{2}\) reverses each string, so it swaps \(by\) and \(yb\) while keeping \(bb\) and \(yy\) fixed.
By Burnside's lemma, the number of equivalence classes equals the average number of elements fixed by the permutations of the group: \(N = \frac{1}{|G|}\sum_{\pi \in G} \psi(\pi)\).
The identity fixes all four strings, so \(\psi(\Pi_{1}) = 4\). The reversal fixes only \(bb\) and \(yy\), so \(\psi(\Pi_{2}) = 2\).
Therefore \(N = \frac{4 + 2}{2} = 3\).
Check: listing the classes directly gives \(\{bb\}\), \(\{by, yb\}\) and \(\{yy\}\), which is exactly 3 distinct strings of beads.
Hence, 3.
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