The question I have been given is as follows:
Recall that each permutation can be written as a product of disjoint cycles. How many permutations of {1, 2, . . . , 8} are a disjoint product of one 1-cycle, two 2-cycles and one 3-cycle.
by doing some research on Wikipedia, I found the formula, $$(k-1)! {n \choose r}$$ This gave me the answer, $$(3-1)!{8 \choose 3}\cdot (2-1)!{5 \choose 2}\cdot(2-1)!{5-2 \choose 2}=(2\cdot56)(1\cdot10)(1\cdot3)=3360$$
This seems correct to me. The problem is that we haven't actually learned that formula in class. Is there another way to logically get the answer without that formula? Is my answer even correct? I would appreciate any help.
This is unfortunately incorrect since the permutation (1 2 3)(4 5)(6 7) (expressed in its disjoint cyclic decomposition form) is equal to the permutation (1 2 3)(6 7)(4 5). You need to be a bit more careful how you handle counting your two 2-cycles. Your attempt distinguished between the "first" 2-cycle and the "second" 2-cycle when they should have been indistinguishable.
As for why are there $\binom{n}{r}(r-1)!$ different $r$-cycles able to be formed out of $n$ available elements? Choose the $r$ elements involved in the cycle first, then place the smallest element at the front as is required for the canonical representation to avoid ambiguity, and then order the remaining $r-1$ elements within the cycle.
To correct your count, you can divide your answer by $2$ since that is how many times you counted each result. Alternatively, you could have counted this directly by adjusting the step where you count the two 2-cycles to have said "First choose the four elements involved in the two 2-cycles. The smallest of those chosen can go at the front of the first of these. Pick which of the others chosen follows. The two remaining then form their own cycle, again with the smallest in the front to avoid ambiguity and to conform to the canonical representation" giving a total of $\binom{5}{4}\cdot 3$ ways to complete this step for a final total of $\binom{8}{3}\cdot 2! \cdot \binom{5}{4}\cdot 3$
Again, in order to avoid counting the same permutation multiple times with multiple different representations for each, the canonical representation of a permutation in disjoint cyclic form will be such that within each cycle... the smallest element of that cycle appears at the front, and then ordering of the cycles is done by putting the leading entries of the cycle in increasing order.