Permutation Cycles is the habit of decomposing permutations into cycles to understand arrangements, parity, and repeated moves. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.
$$\text{Every permutation decomposes uniquely into disjoint cycles.}$$
Follow where each element is sent until you loop back; a permutation is a product of these disjoint cycles.
Why It Works
1
Start at any element and repeatedly apply the permutation; since the set is finite, you must eventually return to the start, closing a cycle.
2
Elements not yet visited start new cycles, and no element appears in two cycles — so the cycles are disjoint and partition the elements.
3
A cycle of length $\ell$ is a single cyclic shift; the whole permutation is the product of its disjoint cycles, and this decomposition is unique up to order.
4
Cycle structure controls order (the lcm of cycle lengths), parity (a length-$\ell$ cycle is $\ell - 1$ transpositions), and counts of permutations with given shape.
Worked Examples
Example 1
How many permutations of $\{1, 2, 3, 4, 5\}$ consist of a single $5$-cycle?
1
Build the cycle by listing its elements in order, starting (without loss of generality) at $1$ to avoid counting the $5$ rotations of the same cycle separately.
2
The remaining four positions of the cycle can be filled by the other elements $\{2,3,4,5\}$ in any order: $4!$ ways.
3
Compute: $4! = 24$.
4
Check: total cyclic arrangements of $5$ items is $\tfrac{5!}{5} = \tfrac{120}{5} = 24$, matching.
Answer:
$24$
Warm-up
A permutation of $\{1, 2, \ldots, 7\}$ is built from a $4$-cycle and a $3$-cycle (cycle type $(4, 3)$). What is its order — the smallest $k$ with the permutation applied $k$ times equal to the identity?
1
Applying the permutation $k$ times returns each element home iff $k$ is a multiple of every cycle length.
2
So the order is the least common multiple of the cycle lengths.
3
Here the lengths are $4$ and $3$, so the order is $\operatorname{lcm}(4, 3) = 12$.
Answer:
$12$
Contest level
How many permutations of $\{1, 2, 3, 4, 5, 6\}$ have cycle type $(3, 3)$ — that is, decompose into two disjoint $3$-cycles?
1
First count ORDERED ways to form two $3$-cycles. Split the $6$ elements into an ordered pair of $3$-element groups: $\binom{6}{3} = 20$ ways to choose the first group (the second is determined).
2
Each group of $3$ forms a $3$-cycle in $(3-1)! = 2$ ways (cyclic arrangements of $3$ items).
3
Ordered count: $20 \cdot 2 \cdot 2 = 80$. But the two $3$-cycles are indistinguishable, so each unordered permutation was counted $2!$ times.
4
Divide by $2$: $\dfrac{80}{2} = 40$.
Answer:
$40$
Olympiad / Challenge
How many permutations of $\{1, 2, \ldots, n\}$ consist of a single $n$-cycle, and what fraction of all $n!$ permutations is that?
1
Build the single $n$-cycle by writing its elements in cyclic order. Fixing the starting element (say always begin the cycle notation at the smallest element) removes the $n$ rotations that describe the same cycle.
2
The remaining $n - 1$ positions are filled by the other $n - 1$ elements in any order: $(n-1)!$ ways.
3
So there are $(n-1)!$ single $n$-cycles. As a fraction of all permutations: $\dfrac{(n-1)!}{n!} = \dfrac{1}{n}$.
4
For example $n = 5$: $4! = 24$ single $5$-cycles, which is $\tfrac{24}{120} = \tfrac{1}{5}$ of all permutations.
Answer:
$(n-1)!$ such permutations, exactly $\dfrac{1}{n}$ of all $n!$.
Going Deeper
The number of permutations of $n$ with $k$ cycles is the unsigned Stirling number of the first kind $\left[{n \atop k}\right]$, and the number with a prescribed cycle type having $c_j$ cycles of length $j$ is $\dfrac{n!}{\prod_j j^{c_j}\,c_j!}$. Each divisor cancels an overcount in the $n!$ ways of writing the elements into the cycle slots: the $j^{c_j}$ divides out the $j$ cyclic rotations of each of the $c_j$ cycles of length $j$, and the $c_j!$ divides out the orderings of those $c_j$ equal-length-$j$ cycles among themselves. Cycle structure also fixes the order (lcm of lengths) and sign ($(-1)^{n - (\#\text{cycles})}$).
Cycle decomposition is central to counting permutations by symmetry (Burnside / Pólya enumeration), to derangements (permutations with no $1$-cycle), and to AIME problems about the order of a shuffle or the number of permutations of a given shape. The 'fix the start to kill rotations' trick recurs whenever you count cyclic structures.
Pitfall: when counting permutations of a given cycle type, you must divide out BOTH the rotations inside each cycle (a $j$-cycle has $j$ rotational descriptions) AND the permutations of equal-length cycles among themselves ($c_j!$). Forgetting the $c_j!$ factor for repeated cycle lengths (as in the $(3,3)$ case) is the classic overcount.
Spot the Signal
Look for problems where the key step is decomposing permutations into cycles to understand arrangements, parity, and repeated moves.
You can describe the hard part as decomposing permutations into cycles to understand arrangements, parity, and repeated moves, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.
Learn the Move
Start by identify the counting object that calls for permutation cycles, then rewrite the givens around it.
Name the relation that makes Permutation Cycles legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.
Avoid These Traps
Do not use Permutation Cycles just because the surface looks familiar; verify the required condition first.
Applying Permutation Cycles because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.
MathGrit Coach Note
Let permutation cycles reveal the counting structure; then compute only what remains.
Try it on:
Practice a contest problem where the key step is decomposing permutations into cycles to understand arrangements, parity, and repeated moves.