I am struggling to understand why, for the cycle shape $(3,1,1)$ in the symmetric group $S_5$, there are $20$ elements.
In my head there should only be $10$, as once you have chosen the 3 elements to be in the 3-cycle the two 1-cycles should choose themselves shouldn't they? Which would be $\binom{5}{3} = 10$.
I would appreciate anyone explaining why you double the answer.
$(123)(4)(5)$ is not the same as $(321)(4)(5)$.
You need to double your count because there are two distinct three cycles containing any three numbers (they are inverse to each other).