Question 481474: pls help me solve this problem in probability... thank you..
a ring of keys can hold 7 keys. In how many ways can the keys be arranged if the ring can be turned over?
Found 2 solutions by Theo, ikleyn:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! ring of keys can hold 7 keys.
the keys are a,b,c,d,e,f,g
the possible number of arrangements would be 7! = 7*6*5*4*3*2*1 = 5040.
let's see what happens if there are only 3 keys.
the number of arrangements would be 3! = 3*2 = 6
those arrangements would be:
a,b,c
a,c,b
b,a,c
b,c,a
c,a,b
c,b,a
if you turn the ring over, then:
a,b,c becomes c,b,a
a,c,b becomes b,c,a
b,a,c becomes c,a,b
b,c,a becomes a,c,b
c,a,b becomes b,a,c
c,b,a becomes a,b,c
since all these arrangements are already accounted for, there are no new arrangements that are created by flipping the ring over.
i would say that you get the same number of ways the keys can be arranged whether or not the ring can be turned over.
Answer by ikleyn(52864)  (Show Source):
You can put this solution on YOUR website! .
.
pls help me solve this problem in probability... thank you..
a ring of keys can hold 7 keys. In how many ways can the keys be arranged if the ring can be turned over?
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The solution and the reasoning in the post by @Theo are incorrect.
I came to bring a correct solution.
The keys are on the ring - it means that we consider circular permutations of 7 keys.
There are (7-1)! = 6! = 720 circular permutations of 7 keys.
Next, the ring can be turned over. It means that we identify arrangements
that differ by the mirror reflection.
So, we should divide 6! = 720 by 2.
The final ANSWER is: the number of all distinguishable arrangements is 720/2 = 360.
Solved correctly.
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The post-solution note
This problem is not from probability, as you mistakenly write in your post.
The area of Math, to which this problem does relate, is called Combinatorics.
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