Suppose we have $4$ identical red beads and $3$ identical blue beads. In how many ways can we form a necklace out of these?
I am a little confused here. Suppose we fix a red bead and treat the problem now as a linear arrangement, then the number of ways to form a necklace would be $\frac{6!}{2.3!.3!}$.
Suppose we fix a blue bead then the number of ways would be $\frac{6!}{2.2!.4!}$ .
What am I doing wrong here?
Also how can we generalize this problem? If we have $n$ beads out of which $r_1$ are of color $a_1$, $r_2$ of color $a_2$ and so on such that $r_1 + r_2 + r_3 + ... + r_k = n$ ,what is the number of ways to form a necklace?.
Note-Here I am assuming clockwise and anticlockwise arrangements to be the same.


If this would not be a necklace then the number of ways to order them would be $\frac{7!}{3!4!}$. But now we may rotate and we have $7$ kind of rotations. So the total becomes $\frac{7!}{3!4!7}=\frac{6!}{3!4!}$.

With the same reasoning we get that in general we have $\frac{(n-1)!}{r_1!r_2!...r_k!}$

  • $\begingroup$ We would divide your answer by 2 because clockwise and anticlockwise are the same. Ain't it? $\endgroup$ – idpd15 Mar 10 '14 at 11:20
  • $\begingroup$ Notice that clockwise turning $k$ times is the same as anti clockwise turning $n-k$ times. $\endgroup$ – Kaladin Mar 10 '14 at 11:33

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