Decrease of phenotypic or fitness effects of environmental variation through plastic responses. It can reduce immediate selection on other inherited traits.

Environment- or experience-dependent change in animal behaviour expressed by an individual. A change within a lifetime must be distinguished from an evolutionary change in its reaction norm.

Fitness costs associated with maintaining or using a responsive system, distinct from poor performance of a particular expressed phenotype. Comparisons should match the expressed outcome while varying the capacity for response.

Environment-dependent variation in developmental trajectories within a genotype. Plasticity need not be adaptive, and an environmentally induced developmental change need not be reversible.

A commonly used epoxidation reagent; its carbons are not incorporated into the alkene-derived epoxide.

For a permutation $p_0,\dots ,p_{n-1}$ of $n$ ordered distinct values, its Lehmer code has entries $c_i=\#\{j>i:p_j<p_i\}$. Its zero-based lexicographic order rank is ${\sum }_{i=0}^{n-1}{c}_i(n-1-i)!$. Each entry chooses among the still unused values; the factorial weights count all completions with a smaller choice at the first differing position.

The unsigned Stirling numbers of the first kind $c(n,k)$ count permutations of $n$ labels having exactly $k$ disjoint permutation cycles, including fixed points as one-cycles. Their signed version is $s(n,k)=(-1{)}^{n-k}c(n,k)$. Equivalently,The sign convention must be stated because both versions are standard. For three labels, the unsigned counts for $k=1,2,3$ are $2,3,1$: two three-cycles, three transpositions with one fixed point, and the identity permutation. This cycle count differs from the Stirling number of the second kind, which counts unordered nonempty blocks without a cyclic order inside each block.

With the descent convention, the Eulerian number $A(n,k)$ counts permutations of $1,\dots ,n$ with exactly $k$ positions $i$ satisfying $a_i>a_{i+1}$. These positions are permutation descents. For $n=3$, the row is $(1,4,1)$: only 123 has no descent and only 321 has two. Inserting the largest label giveswith $A(0,0)=1$ and out-of-range entries zero. Of the $n$ insertion gaps, $k+1$ preserve $k$ descents, while the remaining gaps increase the descent count. Some sources shift the second index by one, so the convention must accompany the symbol.^[ref]

A signed permutation on $n$ labels is a permutation of the symbols $\{\pm 1,\dots ,\pm n\}$ satisfying $w(-i)=-w(i)$. It is specified by the list $w(1),\dots ,w(n)$, whose absolute values form a permutation of $1,\dots ,n$. For example, $(-2,3,-1)$ specifies $w(1)=-2$, $w(2)=3$, $w(3)=-1$. There are $2^{n}n!$ choices: select an ordinary permutation and an independent sign for each entry. Under composition they form the hyperoctahedral group, acting on a hypercube by permuting coordinate axes and reversing their signs.

A rearrangement of the numerical digits of a numeral. Two numerals have the same digit permutation class exactly when their digit multisets agree. If they represent numbers of the same length, a leading zero must be disallowed.

The order of a permutation $\pi$ is the least positive integer $m$ for which ${\pi }^m$ is the identity permutation. If its disjoint permutation cycles have lengths ${\ell }_1,\dots ,{\ell }_r$, thenIndeed a point in an $\ell$-cycle returns exactly after a multiple of $\ell$ steps, so all points return exactly at common multiples. The permutation $(123)(45)$ has order six, while its five moved labels do not by themselves determine its order. The identity permutation has order one.

The Euler zigzag number $A_n$ counts permutations $a_1\cdots a_n$ satisfying $a_1<a_2>a_3<a_4>\cdots$. Reversing every inequality gives the same count by replacing each label $a$ with $n+1-a$. The first values, starting at $n=0$, are $1,1,1,2,5,16$. For $n=3$ the two upward-starting examples are 132 and 231. Their exponential generating function isThis convention fixes the direction of the first inequality; adding both directions for $n\ge 2$ doubles the count.^[ref]

The Steinhaus-Johnson-Trotter algorithm lists all $n!$ permutations so consecutive entries differ by one adjacent transposition. Start at increasing order with an arrow pointing left on each label. A label is mobile if its arrow points to an adjacent smaller label. Swap the largest mobile label in its arrow direction, then reverse the arrows of every larger label; stop when none is mobile. For three labels this gives 123, 132, 312, 321, 231, 213. Recursively, the largest label sweeps across each smaller-label arrangement in alternating directions, visiting every insertion position once. This gives a Gray code on permutations, useful when an objective can be updated cheaply after a local exchange.

An adjacent transposition exchanges consecutive positions $i$ and $i+1$ in a permutation, leaving the others fixed; as a permutation of positions it is $(ii+1)$. Every permutation can be sorted using adjacent transpositions. Swapping an out-of-order adjacent pair decreases its inversion number by one because its comparisons with every third entry have the same total before and after the swap. Consequently the minimum number of adjacent transpositions needed to sort a permutation is its inversion number. For 312, the route 312, 132, 123 uses two swaps.