Answer:
OK! So this is a case of circular permutation. While dealing with this type of problems, we consider the circle to be placed in free space i.e it is symmetric when viewed from and point around it. So to break the symmetry we first fix one of the beads.
Here I am taking a red bead. Now we are left with 4 red and 2 blue beads.
6 beads can be placed in 6! Ways as 1st place will have 6 possibilities, 2nd place will have 5 possibilities and so on which gives
6*5*4*3*2*1 = 6!
But this time we have considered some extra cases as well because the red and blue beads are identical so which red bead has occupied a specific position will not at all matter. So we divide the total by 5! For identical red beads (the fixed red bead will also affect the position of other beads) and 2! For identical blue beads.
Finally we have 6!/(5! *2!) = 720/(120*2) = 3.
So we can have 3 different necklaces.
As the number is small, we can even enlist the possibilities…
B B R R R R R
B R B R R R R
B R R B R R R
