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These permutations have no cycle of length two, while fixed points and longer cycles are allowed. Inclusion-exclusion over specified two-cycles gives the count shown: only disjoint specified pairs can coexist. Their fraction among all permutations tends to $e^{-1/2}$. This condition differs from being a derangement and from refusing to express a permutation as a product of transpositions.

Over a field of characteristic zero, degree-$n$ symmetric tensors are spanned by pure powers $a^{\otimes n}$. Inclusion-exclusion expands ${\sum }_{J\subseteq [n]}(-1{)}^{n-|J|}({\sum }_{j\in J}{a}_j{)}^{\otimes n}$ into the sum of all permutation orderings of $a_1\otimes \cdots \otimes a_n$. Every surviving term uses each of the $n$ labels once. This converts an invariant-tensor spanning problem into pure tensor powers.

Suppose an integral kernel reproduces under convolution and has integrable diagonal with integral $r$, and the displayed single-variable integrals exist. In the determinant $D_n=\det [K(x_i,x_j)]$, permutations fixing $n$ contribute $r{D}_{n-1}$ after integration. For every permutation on $n-1$ letters there are $n-1$ ways to insert $n$ into a cycle; convolution contracts the two adjacent factors, and insertion reverses the permutation sign. These terms contribute $-(n-1)D_{n-1}$. This proves the formula without a symmetry assumption on the kernel.

Insert the successive letters of a permutation by row insertion to obtain an insertion tableau $P$, recording the insertion times in the new boxes of $Q$. This is a bijection between permutations and pairs of standard Young tableaux of a common shape. Reverse insertion removes the box marked by the latest recording time. The first row of $P$ has length equal to the longest increasing subsequence.

The orientation-preserving symmetries of a cube are the $3\times 3$ signed permutation matrices with determinant one. There are $3!2^2=24$ of them. Their group action on the four body diagonals identifies this group with the symmetric group $S_4$. Indeed choose diagonal directions $(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)$, whose sum is zero and whose only linear relation has all coefficients equal. A rotation fixing every diagonal sends each direction to itself or its negative; the relation forces all four signs equal. All negative would give $-I$, which has determinant $-1$, so the kernel of a group homomorphism is trivial. Both groups have order 24, giving the identification. Its face permutations have cycle types $1^6$, $1^{2}4$, $1^2{2}^2$, $2^3$ and $3^2$, occurring $1,6,3,6,8$ times respectively.

A symmetric tensor is unchanged by every permutation of its slots. A symmetric second-rank tensor is the rank-two case; symmetry for higher rank requires invariance under every such slot exchange.

Right multiplication by $e_t=h_t/H_{\lambda }$ on the group algebra is an idempotent with image $A{e}_t$. Every diagonal entry in the permutation basis is the coefficient of the identity in $e_t$, namely $1/H_{\lambda }$. Thus its rank equals its trace, giving $\dim S^{\lambda }=n!/H_{\lambda }$. Triangular standard-tableau ideals and the Robinson–Schensted correspondence identify this dimension with the standard-tableau count.

A transposition is a permutation that exchanges two elements and fixes every other element. Every transposition has order two.