Permutations in a necklace with repeated beads

A count of arrangements cannot be anything but an integer. To get the correct count of necklaces use Burnside's lemma with group $C_{10}$:

• $\binom{10}5=252$ necklaces invariant under identity
• $0$ necklaces invariant under $1/10,3/10,7/10,9/10$ of a turn
• $2$ necklaces invariant under $1/5,2/5,3/5,4/5$ of a turn (alternate the colours)
• $0$ necklaces invariant under $1/2$ of a turn (this would imply that opposite beads are the same colour, hence there are an even number of beads of any colour, which is not the case)

Hence there are $\frac{252+4\cdot2}{10}=26$ possible necklaces; $63$ is also wrong.