We know that the number of permutations of 5 different objects taken all at a time is 5! = 120 when they are arranged in a straight line.

But when the 5 objects will be arranged in a circle, some of the permutations that were considered before will not be considered as valid permutations now. This is because in a circle we can start counting from anywhere and the different arrangements obtained by counting from different objects in any particular arrangement will not be really "different", as they would represent the same circular arrangement of objects overall.

For 5 objects arranged as a, b, c, d and e in a circle, the following all arrangements are the same:

Therefore, the number of circular permutations of 5 different objects will be 5!/5.

If there are 'n' number of different objects, and they are arranged in a circular fashion, then the total number of circular permutations of these 'n' objects is equal to the number of linear permutations of the 'n' objects

Again, this is because

But when the 5 objects will be arranged in a circle, some of the permutations that were considered before will not be considered as valid permutations now. This is because in a circle we can start counting from anywhere and the different arrangements obtained by counting from different objects in any particular arrangement will not be really "different", as they would represent the same circular arrangement of objects overall.

For 5 objects arranged as a, b, c, d and e in a circle, the following all arrangements are the same:

- a, b, c, d, e
- b, c, d, e, a
- c, d, e, a, b
- d, e, a, b, c
- e, a, b, c, d

Therefore, the number of circular permutations of 5 different objects will be 5!/5.

**Now we can derive the formula for circular permutations:**If there are 'n' number of different objects, and they are arranged in a circular fashion, then the total number of circular permutations of these 'n' objects is equal to the number of linear permutations of the 'n' objects

*divided by 'n'*.Circular permutations of 'n' objects = n!/n

= (n(n - 1)! )/n

= (n - 1)!

Again, this is because

__for every other linear permutation that is there, there will exist 'n' such linear permutations that will be considered the same in the light of circular permutations.__
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