Suppose you’re making a necklace of 10 beads, and you have 3 kinds of beads: red, green and blue. How many different necklaces can you make?
How many of these necklaces contain at least 3 red beads? How many are composed of 2 red beads, 3 green beads and 5 blue beads?
At first glance, these may not seem that tricky to answer. After all, consider the similar question: how many ways can you arrange 10 different beads in a necklace? This is easy: count all permutations of 10 beads, 10!, then divide by 20 because we counted each permutation 10 times due to rotation, and counted each of these twice because you can flip the necklace over. Thus the answer is 10!/20 = 181440.
Perhaps we can solve the first problem with a similar technique. Let’s try a smaller case for now and study 6-bead necklaces. The first step is easy: the number of ways to colour 6 beads, where each bead can be red, green or blue, is 36 = 729. Next we put the beads on a necklace, and account for duplicate patterns. For example, RRRRRG and RRRRGR are the same pattern, because you can rotate the necklace to go from one to the other. On the other hand, some patterns such as RRRRRR only get counted once.
One might hope there are only a few different cases so we can tweak the total a little to get the right answer. But the necklace consisting of 5 red beads and 1 green bead appears 6 times, the necklace consisting of all red beads appears once, the necklace consisting of 4 red beads and 2 green beads at opposite ends such as RRGRRG appears 3 times, and so on. We haven’t even considered blue beads yet!
Furthermore, dramatic changes occur when only one more bead is added. For instance, there are 7 colourings equivalent to RRGRRRG, and in fact, each colouring for the 7 bead case will appear either 1, 2, 7, or 14 times.
There’s no easy way out. To answer the question, we must study the symmetry group of the n-bead necklace: the dihedral group of order 2n.
Why doesn’t the symmetry group matter for the easy variant of the question, where each bead is unique? Actually, it does, but the only information we need is the size of the symmetry group. When each bead is unique, different rotation and reflection operations are guaranteed to produce different colourings, so we know we counted each one exactly 20 times.
When beads can have the same colour as other beads, rotations or reflections can leave a colouring unchanged. For example, rotating RRGRRG by 3 beads has no effect. Our counting algorithm must therefore intimately involve the symmetry group.
References
See K. H. Wehrhahn, "Combinatorics: An Introduction".
Answer
Verified
Hint: At first count the number of ways we can choose 5 beads from 8 different beads. Then count in how many ways we can arrange them.Complete step-by-step answer:
It is given in the question that we have 8 different beads.
We have to choose 5 beads from 8 different beads.
We know that the combination is a way of selecting items from a collection, such that the order of selection does not matter.
At first when we are selecting 5 beads, from 8 different beads then the
order of selection does not matter.
So, we can choose 5 beads from 8 different beads in ${}^{8}{{C}_{5}}$ ways.
${}^{8}{{C}_{5}}=\dfrac{8!}{\left( 8-5 \right)!\times 5!}=\dfrac{8!}{3!\times 5!}=\dfrac{5!\times 6\times 7\times 8}{1\times 2\times 3\times 5!}=\dfrac{6\times 7\times 8}{2\times 3}=56$
Therefore, we can select 5 beads from 8 different beads in 56 ways.
Now we have to arrange those 5 beads on a circular ring.
When we will arrange these 5 beads then the order of
arranging them plays a very significant role. If the order changes, that will become a different arrangement. Therefore, we have to use permutation.
Permutation is arranging all the members of a set into some sequence or order, where the order of selection matters.
Now, we can arrange $n$ elements in a circular permutation in $\left( n-1 \right)!$ ways.
So we can arrange 5 beads in a circular ring in $\left( 5-1 \right)=4!=1\times 2\times 3\times 4=24$ ways.
But here we can arrange 5
beads in two directions. One is clockwise, another is anticlockwise. Here in both directions we will get the same arrangement. So, we have to divide 24 by 2.
Therefore the total number of different ways of arranging 5 beads is $\dfrac{24}{2}=12$ .
Now, we can select 5 beads from 8 different beads in 56 ways.
In each way we can arrange them in 12 ways.
So, the total number of ways of selecting 5 beads from 8 different beads and arranging them on a circular ring is $56\times 12=672$
Therefore the total number of ways in which 5 beads, chosen from 8 different beads be threaded on a circular ring is 672.
Note: For circular permutation we have to be careful about the clockwise and anticlockwise directions.
Generally we make a mistake here. If the direction does not matter then we have to divide $\left( n-1 \right)!$ by 2.