A craftsperson has a six different kinds of seashells. How many different bracelets can be constructed if only four shells are to be used in any one bracelet?
So, to clarify some if this still confuses you.
You want to know the number of possible 4 element combinations chosen from a set of 6 distinct elements.
So, k=4 and n=6.
Thus, your solution is given by
$\displaystyle
_{6}C_4=C(6,4) = \binom{6}{4} =\frac{6!}{4!(6-4)!}
$,
whereby "!" denotes the factorial operation.
-Andy
I think the question is a little more subtle:
$\displaystyle {n \choose k} $ is the amount of sets we can compose of length $\displaystyle k$ out of a set of $\displaystyle n$ objects.
(think of it as choosing different colour combinations)
However the question is how many bracelets can we make: Then we should count some permutations too!
Let's denote the set sea-shells {A,B,C,D,E,F}.
We can choose 4 seashells: For example: {A,B,C,D}
A bracelet might look like this for example: *-A-B-C-D-*
(the little * are connected)
This bracelet if ofcourse the same as: *-D-A-B-C-*
But not the same as: *A-C-B-D* while we used the same sea-shells.
For example: we could argue that 2 bracelets are the same if and only if they can be turned into eachother by rotating.
So for every choice of 4 sea-shells we must count the amount of "real " permutations $\displaystyle \sigma \in S_4$.
We need some algebra for that.
Ok. So I hope I'm correct:
We have 4! permutations in the group $\displaystyle S_4$
Observe that the bracelet is invariant under rotating in the 4 symmetry -axis of the bracelet, and invariant under cyclic permutation.
Thus the bracelet is invariant under the permutations of $\displaystyle D_4$
A - B
C - D
Let $\displaystyle \sigma\in S_4$. Observe that for any $\displaystyle \tau_1,\tau_2\in D_4$ we have $\displaystyle \tau_1\sigma\leftrightarrow \tau_2\sigma $.
Since $\displaystyle D_4 \subset S_4$ is a sub-group with 8 elements we have $\displaystyle |S_4/D_4| = 3$.
So the amount of "real different" permutations is 3 and the amount of "real different" bracelets is $\displaystyle 3\cdot {6\choose 4} = 45 $
This is a good point. It was not specified in the problem, but if I had given this a little more thought, I would have reasoned that the set of all possible bracelets is equivalent to the set of all permutations with n=6 and r=4. If I had given it even more thought, I might have also had the sense to eliminate "duplicate bracelets" from the set of permutations. I responded in too much haste to this question. Good pick up. Rep +1 + 1
Hello, roshanhero!
"Bracelet" problems are always tricky.
I've only started on the solution . . .
NOTE: It did not say that four different types of shells are used.A craftsperson has a six different types of seashells.
How many different bracelets can be made if four shells are used in any one bracelet?
. . . . . This further complicates the problem.
Case 1
Four different shells: $\displaystyle WXYZ$
There are: $\displaystyle _6C_4 \:=\:15$ choices for the types of shells.
The four shells can be placed in a circle in: .$\displaystyle 3! \,=\,6$ ways.
But a bracelet can be turned over.
That is: .$\displaystyle \begin{array}{ccc}&W& \\ Z&&X \\ &Y \end{array}$ . is the same bracelet as: .$\displaystyle \begin{array}{ccc}&W& \\ X&&Z\\ &Y& \end{array}$
Therefore, there are: .$\displaystyle \frac{15\cdot6}{2} \:=\:45$ different bracelets
. . which have four different types of shells.
I'll let you work out the other cases . . .
Soroban is right.
Even I did assume that 4 different shells were used but it becomes even more complicated considering we don't have to use different shells.
This is actually quite a difficult problem:
For every Choice of sea-shells we must figure out under wich permutations in $\displaystyle S_4$ the bracelet is invariant.
I think without Algebra and the formula of William Burnside this problem is difficult.
Ok, so this is how we do it:
the number we must calculate is:
number different bracelets = $\displaystyle {6 \choose 1}k_1+ {6\choose 2}k_2+{6\choose 3}k_3+{6\choose 4}k_4$
Where $\displaystyle k_i$ is the number of different ways we can permutate the shells such that it can not be obtained by rotation.
We allready know $\displaystyle k_1 = 1, k_4 = 3$ and we have $\displaystyle k_2 = 2$. Can you figure out $\displaystyle k_3$?
(I calculated 101 different bracelets)