I'm trying to calculate the size of the conjugacy class of cycle type $[1,2,3]$ in $s_6$.
My calculations are as follows:
$$size = \frac{6\times 5\times 4}{3}\times \frac{\frac{3\times 2}{2}}{2}=60$$
My reasoning is that you have $6\times 4\times 3$ $3$-cycles which /3 give the number of distinct $3$-cycles. Similar reasoning for $2$-cycles and then the last division by two is to account for swapping the two and 3 cycle around.
The actual answer is $120$ but I'm a little confused on where my reasoning has gone wrong.
so in a cycle you fix one of the elements say the lowest that you have picked and then permute the other ones. So Actually you dont have to divide by $2$ the $(3x2)/2.$
In a more organized way to see it, fix the fixed point in $6$ ways, from the $5$ remainder pick $2$ and from the $3$ remainder multiply just the order of the greater $2$ to get a total of $$6\binom{5}{2}2=12\binom{5}{2}=120.$$