# Permutations and Combinations

How many ways are there to find the number of permutations of the elements in $\;\{1,2, 3, 4, 5\}\;$ such that it has one cycle of length $2$ and one cycle of length $3$?

Please be as descriptive as possible.

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There are $\binom{5}{3}$ ways to choose three elements for a cycle of length 3, then $2! = 2$ ways that each of triplets of elements can be permuted, e.g., from the triplet $\{1, 2, 3)\},$ two distinct 3-cycles can be formed $(123)$ and $(132)$, each permutation different from the other. So far, that gives us $\binom 53 \cdot 2$ ways a three cycle can be formed.
Then, for the remaining $2$ elements....How many ways are are there to choose $2$ elements from $2$ elements? Only one way: and this is the remaining 2-cycle, whose permutation is itself.
That gives us $$\binom{5}{3} \cdot 2 = \binom{5}{2}\cdot 2 = \dfrac{5!}{3!2!}\cdot 2=\frac{5 \cdot 4}{2} \cdot 2 = 20$$ total ways to form a permutation of 5 elements into a disjoint product of a three-cycle and a two-cycle.
Another popular way of counting this is as $$\frac{5 \cdot 4}{2} \cdot \frac{3 \cdot 2 \cdot 1}{3}.$$ Here you first choose the two elements of the 2-cycle (in order) in $5 \cdot 4$ ways, then divide by 2 as $(a b) = (b a)$. Then you choose the (remaining) three elements for the 3-cycle (in order) in $3 \cdot 2 \cdot 1$ ways, and divide by 3, as $(abc) = (bca) = (cab)$.