The number of distinct bracelets of five beads made up of $$\color{red}{\text{red}}$$, $$\color{Blue}{\text{blue}}$$ and $$\color{green}{\text{green}}$$ beads (two bracelets are indistinguishable if the rotation of one yield another) is:

1. $$243$$
2. $$81$$
3. $$51$$
4. $$47$$

My attempt:

Somewhere it explained as : Combination of 5 beads same color =RRRRR,BBBBB,GGGGG

4 same colors = RRRRB ,RRRRG (similarly Blue and green colors ) =(5!/4! +5!/4!) ⨉3

3 same colors =RRRGB, RRRGG,RRRBB (Similarly other colors)=(5!/3! + 5!/3!2! + 5!/3!2!)⨉3

2 -2 same color , RRBBG, RRGGB, BBGGR =5!/2!2! ⨉3

But, official key is given $$(3)\space 51$$.

Let us consider this problem from a graph theory perspective.

Consider the bracelets made up of five beads, which can be of red, blue or green colour. Then let us have $D =\{1,2,3,4,5\}$ is the set of places the beads can take, $C =\{r, b, g\}$ is the set of colours that a bead in a particular place can take and $C_5$ is the cyclic group of order $5$ generated by the permutation $(12345)$. Then the question is: how many orbits does $C_5$ have?. For the identity, $1\in C_5$, clearly, $F(1) = X$.

For a permutation $\alpha \in C_5$, we denote by $F(\alpha)$ the set of elements fixed by $\alpha$ as $F(\alpha) = \{x\in X: \alpha x=x\}$ where $X$ is considered to be the set of all possible bracelets (which are $3^5 = 243$ in number).

Now, we use the Cauchy-Frobenius lemma to find the number of orbits. The original form of the lemma says that, $$N(\tau) =\frac{1}{|\tau|} \sum_{ \alpha \in \tau} |F (\alpha)|$$ where $N(\tau)$ is the number of orbits.

Returning to our problem, we know that for every non-trivial rotation $\alpha \in C_5$, only the three monochromatic bracelets are invariant under $\alpha$, so $$N(\tau) = \frac{1}{5}[243 + 3 + 3 + 3 + 3] = 51$$ Hope it helps.

Reference: Modern Graph Theory by Bella Bollobas (pg.$277-279$)

• Thanks for nice explanation. Dec 26, 2016 at 9:57

Verify that any necklace may be characterized via one of the patterns below.

Recall that there are $\frac{3!}{(3-n)!}$ ways to put 3 things into n ordered slots.

xxxxx: 3 = $\frac{3!}{(3-1)!}$

xxxxy: 6 = $\frac{3!}{(3-2)!}$

xxxyy: 6 = $\frac{3!}{(3-2)!}$

xxxyz: 6 = $\frac{3!}{(3-3)!}$

xxyyz: 6 = $\frac{3!}{(3-3)!}$

xxyxy: 6 = $\frac{3!}{(3-2)!}$

xxyxz: 6 = $\frac{3!}{(3-3)!}$

xxyzy: 6 = $\frac{3!}{(3-3)!}$

xyzyz: 6 = $\frac{3!}{(3-2)!}$

3 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 = 51

• Thanks for nice explanation. Dec 26, 2016 at 9:57