Math, asked by Riyazkhan1775, 1 year ago

In how many ways can a party of 7 persons arrange themselves around a circular table?

Answers

Answered by Neerajchauhan1
0
720 ways
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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