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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.

Thank you in advance.

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up vote 6 down vote accepted

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.

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I had thought of a similar answer where I first chose the number of ways to select 2 elements for a cycle (5 choose 2) then for the remaining 3-element cycle I would fix the first element and have two ways of the choosing the second element and 1 way of choosing the last element i.e. (1*2*1) ... resulting in (5 choose 2)*2. It is another way of solving this which results in the same answer. – Sarah Feb 5 '13 at 19:24

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 $(abc) = (bca) = (cab)$.

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