Found problems: 3
2023 Taiwan TST Round 1, 4
Let $k$ be a positive integer, and set $n=2^k$, $N=\{1, 2, \cdots, n\}$. For any bijective function $f:N\rightarrow N$, if a set $A\subset N$ contains an element $a\in A$ such that $\{a, f(a), f(f(a)), \cdots\} = A$, then we call $A$ as a cycle of $f$. Prove that: among all bijective functions $f:N\rightarrow N$, at least $\frac{n!}{2}$ of them have number of cycles less than or equal to $2k-1$.
[i]Note: A function is bijective if and only if it is injective and surjective; in other words, it is 1-1 and onto.[/i]
[i]Proposed by CSJL[/i]
2013 India IMO Training Camp, 1
Let $n \ge 2$ be an integer. There are $n$ beads numbered $1, 2, \ldots, n$. Two necklaces made out of some of these beads are considered the same if we can get one by rotating the other (with no flipping allowed). For example, with $n \ge 5$, the necklace with four beads $1, 5, 3, 2$ in the clockwise order is same as the one with $5, 3, 2, 1$ in the clockwise order, but is different from the one with $1, 2, 3, 5$ in the clockwise order.
We denote by $D_0(n)$ (respectively $D_1(n)$) the number of ways in which we can use all the beads to make an even number (resp. an odd number) of necklaces each of length at least $3$. Prove that $n - 1$ divides $D_1(n) - D_0(n)$.
2013 India IMO Training Camp, 1
Let $n \ge 2$ be an integer. There are $n$ beads numbered $1, 2, \ldots, n$. Two necklaces made out of some of these beads are considered the same if we can get one by rotating the other (with no flipping allowed). For example, with $n \ge 5$, the necklace with four beads $1, 5, 3, 2$ in the clockwise order is same as the one with $5, 3, 2, 1$ in the clockwise order, but is different from the one with $1, 2, 3, 5$ in the clockwise order.
We denote by $D_0(n)$ (respectively $D_1(n)$) the number of ways in which we can use all the beads to make an even number (resp. an odd number) of necklaces each of length at least $3$. Prove that $n - 1$ divides $D_1(n) - D_0(n)$.