That'll be a colourful necklace

By the lemma that is not Burnside's , the number of orbits (i.e. different necklaces) under the action of the dihedral group $D_{41}$ (which has 41 rotations + 41 symmetries=82 elements) is:

$N= \displaystyle \frac{1}{|D_{41}|}\sum_{g\in D_{41}} Fix(g)$

The only element of $D_{41}$ which fixes some beads is the identity, because there are no two beads of the same colour, and there are $41!$ arrangements of the beads, so the answer is:

$\frac{1\cdot 41!+40\cdot 0+41\cdot 0}{82}=\frac{41!}{82}=\boxed{4.080E47}$

(n-1)!/2

n=> no of beads to form a necklace

Number of possible different necklaces =( (41!) / ( 41 ) ) / 2