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  "body": "\n## Definition\n\nLet $K$ be a [[commutative ring|0310REJJY8]], and let $A = (a_{ij})$ be an $n \\times n$ [[square matrix|0JXZKX526R]] over $K$.\nThe *permanent* of $A$ is\n$$\\operatorname{perm}(A) = \\sum_{\\sigma \\in S_n} \\prod_{i=1}^{n} a_{i, \\sigma(i)},$$\nwhere $S_n$ is the [[symmetric group|0515A6D2Y1]] of permutations of $\\set{1, \\dots, n}$.\n\n## Remarks\n\nExample: for $A = \\begin{pmatrix} 1 \u0026 2 \\\\ 3 \u0026 4 \\end{pmatrix}$, the two permutations give $\\operatorname{perm}(A) = 1 \\cdot 4 + 2 \\cdot 3 = 10$, while $\\det A = -2$.\n\nThe permanent is not multiplicative, unlike the [[determinant|00KDZRW4C0]].\nFor $J = \\begin{pmatrix} 1 \u0026 1 \\\\ 1 \u0026 1 \\end{pmatrix}$, $\\operatorname{perm}(J) = 2$, but $J^2 = 2J$ has permanent $8 \\ne 2 \\cdot 2$.\n\nThe permanent is linear in each row and each column, and it does not change when rows or columns are permuted.\nIf $A$ is the $0/1$ biadjacency matrix of a [[bipartite graph|0ZVQKTRKVQ]] with $n$ vertices on each side, then $\\operatorname{perm}(A)$ is the number of [[perfect matchings|0NHK9MFFW9]] of the graph.\nThe permanent of the $n \\times n$ all-ones matrix is $n!$.\n\n## Explanations\n\n### L1 human\n\nThe permanent is computed like the determinant, but every term gets a plus sign.\nThis small change makes it lose most of the good algebraic properties of the determinant.\n",
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