Discussion
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 :
\(\Pi_1 = \begin{pmatrix} bb & by & yb & yy \\ bb & by & yb & yy \end{pmatrix}\)
\(\Pi_2 = \begin{pmatrix} bb & by & yb & yy \\ bb & yb & by & yy \end{pmatrix}\)
Answer: A
There are 3 equivalence classes.
The group has two permutations: the identity \(\Pi_1\), and \(\Pi_2\), which swaps by and yb (reversing the string) and fixes bb and yy. Two strings are equivalent when some permutation in the group maps one to the other.
So the classes are {bb}, {by, yb} and {yy}: three distinct strings once the ends are indistinguishable. Burnside's lemma agrees: \(\Pi_1\) fixes 4 strings and \(\Pi_2\) fixes 2, and \((4 + 2)/2 = 3\).
Four is the count before the equivalence is applied; 5 and 6 are larger than the set itself.
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