Intro to Algebra, Symmetric group # of elements of order 4 in S6

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Homework Statement



How many elements of order 4 are in S6? (symmetric group with order 6)

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The Attempt at a Solution



So, the different forms of elements with order 4 in S6 are

(abcd)(ef), (abcd)

from there I am sunk on how to calculate. I know there are 6! = 720 total elements in S6.
 
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well, first, how many 4-cycles are there?

first, we have to pick 4 elements to permute out of 6. how many ways are there to choose 4 elements out of 6?

let's say we've picked our 4: {a,b,c,d}.

how many distinct 4-cycles can you make on that set (this should be the same number of 4-cycles as we have in S4, right?).

that should help you with the "pure" 4 -cycles.

now, suppose you have (a b c d)(e f).

once we've chosen a,b,c,d (step one in counting the 4-cycles), do we really have any choice as how to choose e and f? and given any two elements of {1,2,3,4,5,6}, say, j and k, how many transpositions can be made from j and k? if you think about this the right way, you don't even have to count the 4+2 cycles.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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