Question

Does a circular permutation of 5 distinct objects always have fewer arrangements than a linear permutation of 5 distinct objects? Explain.

Answer 1

Tricky... I wonder how you're supposed to explain this... algebraically? :D

Answer 2

Because you can just flat out use the formulas...

Answer 3

circular permutations are calculated by Pn = (n-1)!
So:
P5 = (5-1)! = 4! = 4*3*2*1 = 24
inear permutations are I believe , the standard permutation of n:
So:
5= 5*4*3*2*1 = 120

Bydefinition... (n-1) is always going to be smaller than n

Answer 4

^that's algebraic
You can always be conceptual :)
you can think of a linear permutation as a circular permutation that has been 'cut' somewhere. But the difference is, given the same circular permutation, the linear permutation that results is very much affected by WHERE you 'cut' the circle. And since there are 5 of them, there are 5 spaces in which to 'cut', increasing the number of permutations by a factor of 5...
\[\Large 5\cdot (5-1)! = 5!\]