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
,
whereby "!" denotes the factorial operation.
-Andy
I think the question is a little more subtle:
is the amount of sets we can compose of length out of a set of 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 .
We need some algebra for that.
Ok. So I hope I'm correct:
We have 4! permutations in the group
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
A - B
C - D
Let . Observe that for any we have .
Since is a sub-group with 8 elements we have .
So the amount of "real different" permutations is 3 and the amount of "real different" bracelets is
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:
There are: choices for the types of shells.
The four shells can be placed in a circle in: . ways.
But a bracelet can be turned over.
That is: . . is the same bracelet as: .
Therefore, there are: . 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 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 =
Where is the number of different ways we can permutate the shells such that it can not be obtained by rotation.
We allready know and we have . Can you figure out ?
(I calculated 101 different bracelets)