aime26-06

Proposed solution

I need to find the number of functions π: A → A such that π⁶(a) = a for all a ∈ A, where A = {1,2,3,4,5,6}.

First, let me understand what π⁶(a) = a means. This means that applying the function π six times returns us to the original element. In other words, π is a permutation of A with order dividing 6.

The possible cycle structures for permutations of 6 elements where the order divides 6 are:
- Cycles of length 1 (fixed points)
- Cycles of length 2 
- Cycles of length 3
- Cycles of length 6

Since the order of a permutation equals the LCM of its cycle lengths, I need cycle lengths whose LCM divides 6.

The divisors of 6 are: 1, 2, 3, 6.

So valid cycle structures have cycle lengths that are divisors of 6.

Let me enumerate all possible combinations:

Case 1: All cycles of length 1 (identity permutation)
This gives us 1 permutation.

Case 2: One cycle of length 2, rest of length 1
We need to choose 2 elements out of 6 for the 2-cycle.
Number of ways = C(6,2) = 15

Case 3: One cycle of length 3, rest of length 1  
We need to choose 3 elements out of 6 for the 3-cycle.
Number of ways = C(6,3) = 20
For each choice of 3 elements, there are (3-1)! = 2! = 2 ways to arrange them in a cycle.

So total = 20 × 2 = 40

Case 4: One cycle of length 6
Number of ways = (6-1)! = 5! = 120

Case 5: Two cycles of length 3, rest of length 1
We need to partition 6 elements into two groups of 3.
Number of ways = C(6,3)/2 = 20/2 = 10
For each group of 3, there are (3-1)! = 2 ways to arrange in a cycle.
Total = 10 × 2 × 2 = 40

Case 6: One cycle of length 2 and one cycle of length 3, rest of length 1
We need to choose 2 elements for the 2-cycle and 3 elements for the 3-cycle.
Number of ways = C(6,2) × C(4,3) = 15 × 4 = 60
For the 2-cycle: 1 way (since (ab) = (ba))
For the 3-cycle: 2 ways
Total = 60 × 1 × 2 = 120

Case 7: Three cycles of length 2, rest of length 1
We need to partition 6 elements into three pairs.
Number of ways = C(6,2) × C(4,2) × C(2,2)/(3!) = 15 × 6 × 1/6 = 15

Case 8: One cycle of length 2 and one cycle of length 4
But 4 doesn't divide 6, so this is invalid.

Wait, let me reconsider what I mean by "order divides 6". 

Actually, the order of a permutation equals the LCM of its cycle lengths. So if we have cycles of lengths k₁, k₂, ..., kₙ, then the order is LCM(k₁, k₂, ..., kₙ).

The valid cycle lengths are those that divide 6: 1, 2, 3, 6.

Let me reconsider all valid partitions of 6:

1. (1,1,1,1,1,1) - six 1-cycles: 1 way
2. (2,1,1,1,1) - one 2-cycle, four 1-cycles: C(6,2) = 15 ways  
3. (3,1,1,1) - one 3-cycle, three 1-cycles: C(6,3) × 2 = 20 × 2 = 40 ways
4. (6) - one 6-cycle: 5! = 120 ways
5. (3,3) - two 3-cycles: C(6,3)/2 × 2 × 2 = 10 × 4 = 40 ways
6. (2,2,1,1) - two 2-cycles, two 1-cycles: C(6,2) × C(4,2)/2 = 15 × 6/2 = 45 ways
7. (2,3,1) - one 2-cycle, one 3-cycle, one 1-cycle: C(6,2) × C(4,3) = 15 × 4 = 60 ways

Wait, I need to be more careful about counting.

For the case of two 2-cycles and two 1-cycles:
We choose 2 elements for first 2-cycle: C(6,2)
We choose 2 elements from remaining 4 for second 2-cycle: C(4,2)  
But we don't want to double count, so we divide by 2!
Total = C(6,2) × C(4,2)/2! = 15 × 6/2 = 45

For the case of one 2-cycle and one 3-cycle:
We choose 2 elements for 2-cycle: C(6,2)
We choose 3 elements from remaining 4 for 3-cycle: C(4,3)  
But this is wrong - we need to choose 3 elements from 6 total.
Actually: C(6,2) × C(4,3) = 15 × 4 = 60

Let me recount properly:

Case 1: (1,1,1,1,1,1): 1 way
Case 2: (2,1,1,1,1): C(6,2) = 15 ways  
Case 3: (3,1,1,1): C(6,3) × 2 = 20 × 2 = 40 ways
Case 4: (6): 5! = 120 ways
Case 5: (3,3): C(6,3)/2 × 2 × 2 = 10 × 4 = 40 ways  
Case 6: (2,2,1,1): C(6,2) × C(4,2)/2 = 15 × 6/2 = 45 ways
Case 7: (2,3,1): C(6,2) × C(4,3) = 15 × 4 = 60 ways

Total = 1 + 15 + 40 + 120 + 40 + 45 + 60 = 321

Wait, I think I'm overcomplicating this. Let me just compute the number of permutations of {1,2,3,4,5,6} with order dividing 6.

The permutations with order dividing 6 are exactly those whose