The number of transpositions in Symmetric group is $n(n-1)/2$

Question

How could I prove it? Any suggestions will be appreciated. I found this in a book, can I take it as a proof? : ''Given the permutation ( 1 , 2) in Sn, what elements commute with it ? Certainly any permutation leaving both I and 2 fixed does. There are (n - 2) ! such. Also ( 1 , 2) commutes with itself. This way we get 2 (n - 2) ! elements in the group generated by ( 1 , 2) and the (n 2) ! permutations leaving l and 2 fixed. Are there others? There are n(n - l ) /2 transpositions and these are precisely all the conjugates of ( 1 , 2). Thus the conjugate class of ( 1 , 2) has in it n(n - l ) /2 elements. If the order of the normalizer of ( 1 , 2) is r, then, by our counting principle, n(n-1)/2''

Answer 1

Simply chose the two symbols to be used in the transposition, this is ${n\choose 2} = {n(n-1)\over 2}$.