Question 953596: in how many ways can a party of seven persons arrange themselves
a. in a row of seven chairs?
b. around a circular table?
using combinatorial analysis
Found 2 solutions by stanbon, Theo:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! in how many ways can a party of seven persons arrange themselves
a. in a row of seven chairs?
Ans:: 7P7 = 7! = 5040 ways
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b. around a circular table?
Ans: 6! = 720 ways
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Cheers,
Stan H.
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! In a row of 7 chairs, the number of permutations would be 7!.
around a circular table it becomes a little more complicated.
this is because there is no fixed first chair.
it could be any one of the 7.
i would estimate the number of permutations would be 7 * 7!.
but i could be wrong, so we'll have to do a demo to see what happens.
we can see how that works with less permutations to deal with.
7! = 5040 permutations.
3! = 6 permutations.
3! is much easier to work with than 7!, so we'll use that as a demo.
let the seats be labeled a,b,c
in a straight line, the 6 possible permutations would be:
abc
acb
bac
bca
cab
cba
now let's take a look at the circular table.
that first arrangement could be repeated 3 times in all.
first time starting with position 1.
second time starting with position 2.
third time starting with position 3.
you would get:
abc
cab
bca
a started in first position.
a started in second position and then wrapped around.
a started in third position and then wrapped around.
cab and bca are different arrangement.
however, they are covered in the primary set that we got from the straight line arrangement, so there are no new arrangements.
let's take a look at the second arrangement of acb.
wrapping that arrangement around as we did the first, we would get:
acb
bac
cba
we have 2 new arrangements of bac and cba.
bac and cba are, however, already covered in the straight line permutations.
let's jump to the last arrangement of cba.
wrapping them around as we did the first of the circulars, we will get:
cba
acb
bca
we have 2 additional arrangement again.
they are acb and bca.
they are also covered in the initial set of arrangements from the straight line version.
so it appears that the straight line arrangement and the circular arrangement give you the same number of permutations.
i made it even simpler to see if we get the same result.
i assumed 2 arrangements called a and b.
straight line:
there are 2 possible permutations.
they are:
ab
ba
i took ab and then rotated the positions around the table and got:
ab amd ba.
i took ba and then rotated the positions around the table and got:
ba and ab.
i got 4 arrangements but 2 of them were identical so that reduced to 2 distinct arrangements.
the number of permutations was the same whether i used a straight line presentation or a circular presentation.
i'll go with that.
you can do the same experiment for yourself to see if you get the same conclusion.
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