{
  "index": "2024-A-3",
  "type": "COMB",
  "tag": [
    "COMB",
    "ALG"
  ],
  "difficulty": "",
  "question": "Let $S$ be the set of bijections\n\\[\nT \\colon \\{1,2,3\\} \\times \\{1,2,\\dots,2024\\} \\to \\{1,2,\\dots,6072\\}\n\\]\nsuch that $T(1,j) < T(2,j) < T(3,j)$ for all $j \\in \\{1,2,\\dots,2024\\}$ and $T(i,j) < T(i,j+1)$ for all $i \\in \\{1,2,3\\}$ and $j \\in \\{1,2,\\dots,2023\\}$. Do there exist $a$ and $c$ in $\\{1,2,3\\}$ and $b$ and $d$ in $\\{1,2,\\dots,2024\\}$ such that the fraction of elements $T$ in $S$ for which $T(a,b) < T(c,d)$ is at least $1/3$ and at most $2/3$?",
  "solution": "Yes, such $a,b,c,d$ exist: we take\n\\[\n(a,b) = (2,1), \\qquad (c,d) = (1,2).\n\\]\nWe will represent $T$ as an $3 \\times n$ array (3 rows, $n$ columns) of integers  in which each of $1,\\dots,3n$ occurs exactly once and the rows and columns are strictly increasing; we will specialize to $n=2024$ at the end.\n\nWe first note that $T(1,1) = 1$ and $2 \\in \\{T(1,2), T(2,1)\\}$.\nFrom this, it follows that $T(2,1) < T(1,2)$ if and only if $T(2,1) = 2$.\n\nWe next recall a restricted form of the \\emph{hook length formula}\n(see the first remark for a short proof of this restricted version and the second remark for the statement of the general formula).\nConsider more generally an array consisting of (up to) three rows of lengths $n_1\\geq n_2 \\geq n_3 \\geq 0$, aligned at the left.\nLet $f(n_1,n_2,n_3)$ be the number of ways to fill this array with a permutation of the numbers $1,\\dots,n_1+n_2+n_3$ in such a way that each row increases from left to right and each column increases from top to bottom. The hook length formula then shows that $f(n_1,n_2,n_3)$ equals\n\\[\n\\frac{(n_1-n_2+1)(n_1-n_3+2)(n_2-n_3+1) (n_1+n_2+n_3)!}\n{(n_1+2)! (n_2+1)! n_3!}.\n\\]\n\nWe then note that if $T(2,1) = 2$, we obtain a array with row lengths $n, n-1, n-1$ by removing 1 and 2, relabeling each remaining $i$ as $3n+1-i$, and reflecting in both axes. The probability that $T(2,1) < T(1,2)$ is thus\n\\begin{align*}\n\\frac{f(n,n-1,n-1)}{f(n,n,n)}\n&= \n\\frac{(2)(3)(n+1)n}{(1)(2) (3n)(3n-1)} \\\\\n&= \\frac{n+1}{3n-1} = \\frac{1}{3} + \\frac{4}{9n-3};\n\\end{align*}\nthis is always greater than $\\frac{1}{3}$, and for $n =2024$ it is visibly less than $\\frac{2}{3}$.\n\n\\noindent\n\\textbf{Remark.}\nWe prove the claimed formula for $f(n_1,n_2,n_3)$ by induction on $n_1+n_2+n_3$.\nTo begin with, if $n_2 = n_3 = 0$, then the desired count is indeed $f(n_1, 0, 0) = 1$.\nNext, suppose $n_2 > 0, n_3 = 0$.\nThe entry $n_1 + n_2$ must go at the end of either the first or second row;\ncounting ways to complete the diagram from these starting points yields\n\\[\nf(n_1,n_2,0) = f(n_1-1,n_2,0) + f(n_1,n_2-1,0).\n\\]\n(This works even if $n_1 = n_2$, in which case the first row is not an option but correspondingly $f(n_2-1,n_2,0) = 0$.)\nThe induction step then follows from the identity\n\\[\n\\frac{(n_1-n_2)(n_1+1) + (n_1-n_2+2)n_2}{(n_1-n_2+1)(n_1+n_2)}  = 1.\n\\]\n(As an aside, the case $n_1 = n_2, n_3 = 0$ recovers a standard interpretation of the Catalan numbers.)\n\nFinally, suppose $n_3 > 0$. We then have\n\\begin{align*}\n&f(n_1,n_2,n_3) \\\\\n&= \nf(n_1-1,n_2,n_3) + f(n_1,n_2-1,n_3) + f(n_1,n_2,n_3-1),\n\\end{align*}\nand the induction step now reduces to the algebraic identity\n\\begin{align*}\n&\\frac{(n_1-n_2)(n_1-n_3+1)(n_1+2)}{(n_1-n_2+1)(n_1-n_3+2)(n_1+n_2+n_3)} \\\\\n&+ \\frac{(n_1-n_2+2)(n_2-n_3)(n_2+1)}{(n_1-n_2+1)(n_2-n_3+1)(n_1+n_2+n_3)} \\\\\n&+ \\frac{(n_1-n_3+3)(n_2-n_3+2)n_3}{(n_1-n_3+2)(n_2-n_3+1)(n_1+n_2+n_3)}\n= 1.\n\\end{align*}\n\n\\noindent\n\\textbf{Remark.}\nWe formulate the general hook length formula in standard terminology.\nLet $N$ be a positive integer, and consider a semi-infinite checkerboard with top and left edges. A \\emph{Ferrers diagram} is a finite subset of the squares of the board which is closed under taking a unit step towards either edge. Given a Ferrers diagram with $N$ squares, a \\emph{standard Young tableau} for this diagram is a bijection of the squares of the diagram with the integers $1,\\dots,N$ such that the numbers always increase under taking a unit step away from either edge. \n\n\nFor each square $s = (i,j)$ in the diagram, the \\emph{hook length} $h_s$ of $s$ is the number of squares $(i',j')$ in the diagram\nsuch that either $i=i', j\\leq j'$ or $i\\leq i',j=j'$ (including $s$ itself). Then the number of standard Young tableaux for this diagram equals\n\\[\n\\frac{N!}{\\prod_s h_s}.\n\\]\nFor a proof along the lines of the argument given in the previous remark, see:\nKenneth Glass and Chi-Keung Ng, A simple proof of the hook length formula,\n\\textit{American Mathematical Monthly} \\textbf{111} (2004), 700--704.",
  "vars": [
    "S",
    "T",
    "j",
    "i",
    "a",
    "b",
    "c",
    "d",
    "s",
    "f"
  ],
  "params": [
    "n",
    "n_1",
    "n_2",
    "n_3",
    "N",
    "h_s"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "S": "bijectionset",
        "T": "bijectionmap",
        "j": "columnindex",
        "i": "rowindex",
        "a": "rowchoice",
        "b": "colchoice",
        "c": "otherrow",
        "d": "othercolumn",
        "s": "boardsquare",
        "f": "arraycount",
        "n": "columncount",
        "n_1": "lengthfirst",
        "n_2": "lengthsecond",
        "n_3": "lengththird",
        "N": "totalsquares",
        "h_s": "hookmetric"
      },
      "question": "Let $bijectionset$ be the set of bijections\n\\[\nbijectionmap \\colon \\{1,2,3\\} \\times \\{1,2,\\dots,2024\\} \\to \\{1,2,\\dots,6072\\}\n\\]\nsuch that $bijectionmap(1,\\columnindex) < bijectionmap(2,\\columnindex) < bijectionmap(3,\\columnindex)$ for all $\\columnindex \\in \\{1,2,\\dots,2024\\}$ and $bijectionmap(\\rowindex,\\columnindex) < bijectionmap(\\rowindex,\\columnindex+1)$ for all $\\rowindex \\in \\{1,2,3\\}$ and $\\columnindex \\in \\{1,2,\\dots,2023\\}$. Do there exist $\\rowchoice$ and $\\otherrow$ in $\\{1,2,3\\}$ and $\\colchoice$ and $\\othercolumn$ in $\\{1,2,\\dots,2024\\}$ such that the fraction of elements $bijectionmap$ in $bijectionset$ for which $bijectionmap(\\rowchoice,\\colchoice) < bijectionmap(\\otherrow,\\othercolumn)$ is at least $1/3$ and at most $2/3$?",
      "solution": "Yes, such $\\rowchoice,\\colchoice,\\otherrow,\\othercolumn$ exist: we take\n\\[\n(\\rowchoice,\\colchoice) = (2,1), \\qquad (\\otherrow,\\othercolumn) = (1,2).\n\\]\nWe will represent $bijectionmap$ as an $3 \\times \\columncount$ array (3 rows, $\\columncount$ columns) of integers in which each of $1,\\dots,3\\columncount$ occurs exactly once and the rows and columns are strictly increasing; we will specialize to $\\columncount=2024$ at the end.\n\nWe first note that $bijectionmap(1,1) = 1$ and $2 \\in \\{bijectionmap(1,2), bijectionmap(2,1)\\}$. From this, it follows that $bijectionmap(2,1) < bijectionmap(1,2)$ if and only if $bijectionmap(2,1) = 2$.\n\nWe next recall a restricted form of the \\emph{hook length formula} (see the first remark for a short proof of this restricted version and the second remark for the statement of the general formula). Consider more generally an array consisting of (up to) three rows of lengths $\\lengthfirst\\geq \\lengthsecond \\geq \\lengththird \\geq 0$, aligned at the left. Let $\\arraycount(\\lengthfirst,\\lengthsecond,\\lengththird)$ be the number of ways to fill this array with a permutation of the numbers $1,\\dots,\\lengthfirst+\\lengthsecond+\\lengththird$ in such a way that each row increases from left to right and each column increases from top to bottom. The hook length formula then shows that $\\arraycount(\\lengthfirst,\\lengthsecond,\\lengththird)$ equals\n\\[\n\\frac{(\\lengthfirst-\\lengthsecond+1)(\\lengthfirst-\\lengththird+2)(\\lengthsecond-\\lengththird+1) (\\lengthfirst+\\lengthsecond+\\lengththird)!}\n{(\\lengthfirst+2)! (\\lengthsecond+1)! \\lengththird!}.\n\\]\n\nWe then note that if $bijectionmap(2,1) = 2$, we obtain a array with row lengths $\\columncount, \\columncount-1, \\columncount-1$ by removing 1 and 2, relabeling each remaining $\\rowindex$ as $3\\columncount+1-\\rowindex$, and reflecting in both axes. The probability that $bijectionmap(2,1) < bijectionmap(1,2)$ is thus\n\\begin{align*}\n\\frac{\\arraycount(\\columncount,\\columncount-1,\\columncount-1)}{\\arraycount(\\columncount,\\columncount,\\columncount)}\n&= \n\\frac{(2)(3)(\\columncount+1)\\columncount}{(1)(2) (3\\columncount)(3\\columncount-1)} \\\\\n&= \\frac{\\columncount+1}{3\\columncount-1} = \\frac{1}{3} + \\frac{4}{9\\columncount-3};\n\\end{align*}\nthis is always greater than $\\frac{1}{3}$, and for $\\columncount =2024$ it is visibly less than $\\frac{2}{3}$.\n\n\\noindent\n\\textbf{Remark.} We prove the claimed formula for $\\arraycount(\\lengthfirst,\\lengthsecond,\\lengththird)$ by induction on $\\lengthfirst+\\lengthsecond+\\lengththird$. To begin with, if $\\lengthsecond = \\lengththird = 0$, then the desired count is indeed $\\arraycount(\\lengthfirst, 0, 0) = 1$. Next, suppose $\\lengthsecond > 0, \\lengththird = 0$. The entry $\\lengthfirst + \\lengthsecond$ must go at the end of either the first or second row; counting ways to complete the diagram from these starting points yields\n\\[\n\\arraycount(\\lengthfirst,\\lengthsecond,0) = \\arraycount(\\lengthfirst-1,\\lengthsecond,0) + \\arraycount(\\lengthfirst,\\lengthsecond-1,0).\n\\]\n(This works even if $\\lengthfirst = \\lengthsecond$, in which case the first row is not an option but correspondingly $\\arraycount(\\lengthsecond-1,\\lengthsecond,0) = 0$.) The induction step then follows from the identity\n\\[\n\\frac{(\\lengthfirst-\\lengthsecond)(\\lengthfirst+1) + (\\lengthfirst-\\lengthsecond+2)\\lengthsecond}{(\\lengthfirst-\\lengthsecond+1)(\\lengthfirst+\\lengthsecond)}  = 1.\n\\]\n(As an aside, the case $\\lengthfirst = \\lengthsecond, \\lengththird = 0$ recovers a standard interpretation of the Catalan numbers.)\n\nFinally, suppose $\\lengththird > 0$. We then have\n\\begin{align*}\n&\\arraycount(\\lengthfirst,\\lengthsecond,\\lengththird) \\\\\n&= \n\\arraycount(\\lengthfirst-1,\\lengthsecond,\\lengththird) + \\arraycount(\\lengthfirst,\\lengthsecond-1,\\lengththird) + \\arraycount(\\lengthfirst,\\lengthsecond,\\lengththird-1),\n\\end{align*}\nand the induction step now reduces to the algebraic identity\n\\begin{align*}\n&\\frac{(\\lengthfirst-\\lengthsecond)(\\lengthfirst-\\lengththird+1)(\\lengthfirst+2)}{(\\lengthfirst-\\lengthsecond+1)(\\lengthfirst-\\lengththird+2)(\\lengthfirst+\\lengthsecond+\\lengththird)} \\\\\n&+ \\frac{(\\lengthfirst-\\lengthsecond+2)(\\lengthsecond-\\lengththird)(\\lengthsecond+1)}{(\\lengthfirst-\\lengthsecond+1)(\\lengthsecond-\\lengththird+1)(\\lengthfirst+\\lengthsecond+\\lengththird)} \\\\\n&+ \\frac{(\\lengthfirst-\\lengththird+3)(\\lengthsecond-\\lengththird+2)\\lengththird}{(\\lengthfirst-\\lengththird+2)(\\lengthsecond-\\lengththird+1)(\\lengthfirst+\\lengthsecond+\\lengththird)}\n= 1.\n\\end{align*}\n\n\\noindent\n\\textbf{Remark.} We formulate the general hook length formula in standard terminology. Let $\\totalsquares$ be a positive integer, and consider a semi-infinite checkerboard with top and left edges. A \\emph{Ferrers diagram} is a finite subset of the squares of the board which is closed under taking a unit step towards either edge. Given a Ferrers diagram with $\\totalsquares$ squares, a \\emph{standard Young tableau} for this diagram is a bijection of the squares of the diagram with the integers $1,\\dots,\\totalsquares$ such that the numbers always increase under taking a unit step away from either edge. \n\nFor each square $\\boardsquare = (\\rowindex,\\columnindex)$ in the diagram, the \\emph{hook length} $\\hookmetric$ of $\\boardsquare$ is the number of squares $(\\rowindex',\\columnindex')$ in the diagram such that either $\\rowindex=\\rowindex', \\columnindex\\leq \\columnindex'$ or $\\rowindex\\leq \\rowindex',\\columnindex=\\columnindex'$ (including $\\boardsquare$ itself). Then the number of standard Young tableaux for this diagram equals\n\\[\n\\frac{\\totalsquares!}{\\prod_{\\boardsquare} \\hookmetric}.\n\\]\nFor a proof along the lines of the argument given in the previous remark, see: Kenneth Glass and Chi-Keung Ng, A simple proof of the hook length formula, \\textit{American Mathematical Monthly} \\textbf{111} (2004), 700--704."
    },
    "descriptive_long_confusing": {
      "map": {
        "S": "sandstone",
        "T": "trellises",
        "j": "juncture",
        "i": "itinerary",
        "a": "alabaster",
        "b": "backwater",
        "c": "cranberry",
        "d": "daydream",
        "s": "sunflowers",
        "f": "feathered",
        "n": "nightfall",
        "n_1": "marigolds",
        "n_2": "seashores",
        "n_3": "bluebirds",
        "N": "navigation",
        "h_s": "candlestick"
      },
      "question": "Let $sandstone$ be the set of bijections\n\\[\ntrellises \\colon \\{1,2,3\\} \\times \\{1,2,\\dots,2024\\} \\to \\{1,2,\\dots,6072\\}\n\\]\nsuch that $trellises(1,juncture) < trellises(2,juncture) < trellises(3,juncture)$ for all $juncture \\in \\{1,2,\\dots,2024\\}$ and $trellises(itinerary,juncture) < trellises(itinerary,juncture+1)$ for all $itinerary \\in \\{1,2,3\\}$ and $juncture \\in \\{1,2,\\dots,2023\\}$. Do there exist $alabaster$ and $cranberry$ in \\{1,2,3\\} and $backwater$ and $daydream$ in \\{1,2,\\dots,2024\\} such that the fraction of elements $trellises$ in $sandstone$ for which $trellises(alabaster,backwater) < trellises(cranberry,daydream)$ is at least $1/3$ and at most $2/3$?",
      "solution": "Yes, such $alabaster,backwater,cranberry,daydream$ exist: we take\n\\[\n(alabaster,backwater) = (2,1), \\qquad (cranberry,daydream) = (1,2).\n\\]\nWe will represent $trellises$ as an $3 \\times nightfall$ array (3 rows, $nightfall$ columns) of integers  in which each of $1,\\dots,3nightfall$ occurs exactly once and the rows and columns are strictly increasing; we will specialize to $nightfall=2024$ at the end.\n\nWe first note that $trellises(1,1) = 1$ and $2 \\in \\{trellises(1,2), trellises(2,1)\\}$. From this, it follows that $trellises(2,1) < trellises(1,2)$ if and only if $trellises(2,1) = 2$.\n\nWe next recall a restricted form of the \\emph{hook length formula} (see the first remark for a short proof of this restricted version and the second remark for the statement of the general formula). Consider more generally an array consisting of (up to) three rows of lengths $marigolds\\geq seashores \\geq bluebirds \\geq 0$, aligned at the left. Let $feathered(marigolds,seashores,bluebirds)$ be the number of ways to fill this array with a permutation of the numbers $1,\\dots,marigolds+seashores+bluebirds$ in such a way that each row increases from left to right and each column increases from top to bottom. The hook length formula then shows that $feathered(marigolds,seashores,bluebirds)$ equals\n\\[\n\\frac{(marigolds-seashores+1)(marigolds-bluebirds+2)(seashores-bluebirds+1) (marigolds+seashores+bluebirds)!}{(marigolds+2)! (seashores+1)! bluebirds!}.\n\\]\n\nWe then note that if $trellises(2,1) = 2$, we obtain a array with row lengths $nightfall, nightfall-1, nightfall-1$ by removing 1 and 2, relabeling each remaining $itinerary$ as $3nightfall+1-itinerary$, and reflecting in both axes. The probability that $trellises(2,1) < trellises(1,2)$ is thus\n\\begin{align*}\n\\frac{feathered(nightfall,nightfall-1,nightfall-1)}{feathered(nightfall,nightfall,nightfall)}\n&= \\frac{(2)(3)(nightfall+1)nightfall}{(1)(2) (3nightfall)(3nightfall-1)} \\\\\n&= \\frac{nightfall+1}{3nightfall-1} = \\frac{1}{3} + \\frac{4}{9nightfall-3};\n\\end{align*}\nthis is always greater than $\\frac{1}{3}$, and for $nightfall =2024$ it is visibly less than $\\frac{2}{3}$.\n\n\\noindent\\textbf{Remark.} We prove the claimed formula for $feathered(marigolds,seashores,bluebirds)$ by induction on $marigolds+seashores+bluebirds$. To begin with, if $seashores = bluebirds = 0$, then the desired count is indeed $feathered(marigolds, 0, 0) = 1$. Next, suppose $seashores > 0, bluebirds = 0$. The entry $marigolds + seashores$ must go at the end of either the first or second row; counting ways to complete the diagram from these starting points yields\n\\[\nfeathered(marigolds,seashores,0) = feathered(marigolds-1,seashores,0) + feathered(marigolds,seashores-1,0).\n\\]\n(This works even if $marigolds = seashores$, in which case the first row is not an option but correspondingly $feathered(seashores-1,seashores,0) = 0$.) The induction step then follows from the identity\n\\[\n\\frac{(marigolds-seashores)(marigolds+1) + (marigolds-seashores+2)seashores}{(marigolds-seashores+1)(marigolds+seashores)}  = 1.\n\\]\n(As an aside, the case $marigolds = seashores, bluebirds = 0$ recovers a standard interpretation of the Catalan numbers.)\n\nFinally, suppose $bluebirds > 0$. We then have\n\\begin{align*}\n&feathered(marigolds,seashores,bluebirds) \\\\\n&= feathered(marigolds-1,seashores,bluebirds) + feathered(marigolds,seashores-1,bluebirds) + feathered(marigolds,seashores,bluebirds-1),\n\\end{align*}\nand the induction step now reduces to the algebraic identity\n\\begin{align*}\n&\\frac{(marigolds-seashores)(marigolds-bluebirds+1)(marigolds+2)}{(marigolds-seashores+1)(marigolds-bluebirds+2)(marigolds+seashores+bluebirds)} \\\\\n&+ \\frac{(marigolds-seashores+2)(seashores-bluebirds)(seashores+1)}{(marigolds-seashores+1)(seashores-bluebirds+1)(marigolds+seashores+bluebirds)} \\\\\n&+ \\frac{(marigolds-bluebirds+3)(seashores-bluebirds+2)bluebirds}{(marigolds-bluebirds+2)(seashores-bluebirds+1)(marigolds+seashores+bluebirds)} = 1.\n\\end{align*}\n\n\\noindent\\textbf{Remark.} We formulate the general hook length formula in standard terminology. Let $navigation$ be a positive integer, and consider a semi-infinite checkerboard with top and left edges. A \\emph{Ferrers diagram} is a finite subset of the squares of the board which is closed under taking a unit step towards either edge. Given a Ferrers diagram with $navigation$ squares, a \\emph{standard Young tableau} for this diagram is a bijection of the squares of the diagram with the integers $1,\\dots,navigation$ such that the numbers always increase under taking a unit step away from either edge.\n\nFor each square $sunflowers = (itinerary',juncture')$ in the diagram, the \\emph{hook length} $candlestick$ of $sunflowers$ is the number of squares $(itinerary',juncture')$ in the diagram such that either $itinerary=itinerary', juncture\\leq juncture'$ or $itinerary\\leq itinerary',juncture=juncture'$ (including $sunflowers$ itself). Then the number of standard Young tableaux for this diagram equals\n\\[\n\\frac{navigation!}{\\prod_{sunflowers} candlestick}.\n\\]\nFor a proof along the lines of the argument given in the previous remark, see: Kenneth Glass and Chi-Keung Ng, A simple proof of the hook length formula, \\textit{American Mathematical Monthly} \\textbf{111} (2004), 700--704."
    },
    "descriptive_long_misleading": {
      "map": {
        "S": "multiset",
        "T": "nonbiject",
        "j": "constant",
        "i": "columnidx",
        "a": "nonalpha",
        "b": "terminus",
        "c": "beginner",
        "d": "finishers",
        "s": "roundness",
        "f": "emptyness",
        "n": "brevityy",
        "n_1": "tinyfirst",
        "n_2": "tinysecond",
        "n_3": "tinythird",
        "N": "voidsize",
        "h_s": "linegapp"
      },
      "question": "Let $multiset$ be the set of bijections\n\\[\nnonbiject \\colon \\{1,2,3\\} \\times \\{1,2,\\dots,2024\\} \\to \\{1,2,\\dots,6072\\}\n\\]\nsuch that $nonbiject(1,constant) < nonbiject(2,constant) < nonbiject(3,constant)$ for all $constant \\in \\{1,2,\\dots,2024\\}$ and $nonbiject(columnidx,constant) < nonbiject(columnidx,constant+1)$ for all $columnidx \\in \\{1,2,3\\}$ and $constant \\in \\{1,2,\\dots,2023\\}$. Do there exist $nonalpha$ and $beginner$ in \\{1,2,3\\} and $terminus$ and $finishers$ in \\{1,2,\\dots,2024\\} such that the fraction of elements $nonbiject$ in $multiset$ for which $nonbiject(nonalpha,terminus) < nonbiject(beginner,finishers)$ is at least $1/3$ and at most $2/3$?",
      "solution": "Yes, such $nonalpha,terminus,beginner,finishers$ exist: we take\n\\[\n(nonalpha,terminus) = (2,1), \\qquad (beginner,finishers) = (1,2).\n\\]\nWe will represent $nonbiject$ as an $3 \\times brevityy$ array (3 rows, $brevityy$ columns) of integers in which each of $1,\\dots,3brevityy$ occurs exactly once and the rows and columns are strictly increasing; we will specialize to $brevityy=2024$ at the end.\n\nWe first note that $nonbiject(1,1) = 1$ and $2 \\in \\{nonbiject(1,2), nonbiject(2,1)\\}$. From this, it follows that $nonbiject(2,1) < nonbiject(1,2)$ if and only if $nonbiject(2,1) = 2$.\n\nWe next recall a restricted form of the \\emph{hook length formula} (see the first remark for a short proof of this restricted version and the second remark for the statement of the general formula). Consider more generally an array consisting of (up to) three rows of lengths $tinyfirst\\geq tinysecond \\geq tinythird \\geq 0$, aligned at the left. Let $emptyness(tinyfirst,tinysecond,tinythird)$ be the number of ways to fill this array with a permutation of the numbers $1,\\dots,tinyfirst+tinysecond+tinythird$ in such a way that each row increases from left to right and each column increases from top to bottom. The hook length formula then shows that $emptyness(tinyfirst,tinysecond,tinythird)$ equals\n\\[\n\\frac{(tinyfirst-tinysecond+1)(tinyfirst-tinythird+2)(tinysecond-tinythird+1) (tinyfirst+tinysecond+tinythird)!}{(tinyfirst+2)! (tinysecond+1)! tinythird!}.\n\\]\n\nWe then note that if $nonbiject(2,1) = 2$, we obtain an array with row lengths $brevityy, brevityy-1, brevityy-1$ by removing 1 and 2, relabeling each remaining $columnidx$ as $3brevityy+1-columnidx$, and reflecting in both axes. The probability that $nonbiject(2,1) < nonbiject(1,2)$ is thus\n\\begin{align*}\n\\frac{emptyness(brevityy,brevityy-1,brevityy-1)}{emptyness(brevityy,brevityy,brevityy)}\n&= \\frac{(2)(3)(brevityy+1)brevityy}{(1)(2)(3brevityy)(3brevityy-1)} \\\\\n&= \\frac{brevityy+1}{3brevityy-1} = \\frac{1}{3} + \\frac{4}{9brevityy-3};\n\\end{align*}\nthis is always greater than $\\tfrac{1}{3}$, and for $brevityy =2024$ it is visibly less than $\\tfrac{2}{3}$.\n\n\\noindent\\textbf{Remark.}\nWe prove the claimed formula for $emptyness(tinyfirst,tinysecond,tinythird)$ by induction on $tinyfirst+tinysecond+tinythird$. To begin with, if $tinysecond = tinythird = 0$, then the desired count is indeed $emptyness(tinyfirst,0,0) = 1$. Next, suppose $tinysecond > 0, tinythird = 0$. The entry $tinyfirst + tinysecond$ must go at the end of either the first or second row; counting ways to complete the diagram from these starting points yields\n\\[\nemptyness(tinyfirst,tinysecond,0) = emptyness(tinyfirst-1,tinysecond,0) + emptyness(tinyfirst,tinysecond-1,0).\n\\]\n(This works even if $tinyfirst = tinysecond$, in which case the first row is not an option but correspondingly $emptyness(tinysecond-1,tinysecond,0) = 0$.) The induction step then follows from the identity\n\\[\n\\frac{(tinyfirst-tinysecond)(tinyfirst+1) + (tinyfirst-tinysecond+2)tinysecond}{(tinyfirst-tinysecond+1)(tinyfirst+tinysecond)} = 1.\n\\]\n(As an aside, the case $tinyfirst = tinysecond, tinythird = 0$ recovers a standard interpretation of the Catalan numbers.)\n\nFinally, suppose $tinythird > 0$. We then have\n\\begin{align*}\n&emptyness(tinyfirst,tinysecond,tinythird) \\\\\n&= emptyness(tinyfirst-1,tinysecond,tinythird) + emptyness(tinyfirst,tinysecond-1,tinythird) + emptyness(tinyfirst,tinysecond,tinythird-1),\n\\end{align*}\nand the induction step now reduces to the algebraic identity\n\\begin{align*}\n&\\frac{(tinyfirst-tinysecond)(tinyfirst-tinythird+1)(tinyfirst+2)}{(tinyfirst-tinysecond+1)(tinyfirst-tinythird+2)(tinyfirst+tinysecond+tinythird)} \\\\\n&+ \\frac{(tinyfirst-tinysecond+2)(tinysecond-tinythird)(tinysecond+1)}{(tinyfirst-tinysecond+1)(tinysecond-tinythird+1)(tinyfirst+tinysecond+tinythird)} \\\\\n&+ \\frac{(tinyfirst-tinythird+3)(tinysecond-tinythird+2)tinythird}{(tinyfirst-tinythird+2)(tinysecond-tinythird+1)(tinyfirst+tinysecond+tinythird)} = 1.\n\\end{align*}\n\n\\noindent\\textbf{Remark.}\nWe formulate the general hook length formula in standard terminology. Let $voidsize$ be a positive integer, and consider a semi-infinite checkerboard with top and left edges. A \\emph{Ferrers diagram} is a finite subset of the squares of the board which is closed under taking a unit step towards either edge. Given a Ferrers diagram with $voidsize$ squares, a \\emph{standard Young tableau} for this diagram is a bijection of the squares of the diagram with the integers $1,\\dots,voidsize$ such that the numbers always increase under taking a unit step away from either edge.\n\nFor each square $roundness = (columnidx,constant)$ in the diagram, the \\emph{hook length} $linegapp$ of $roundness$ is the number of squares $(columnidx',constant')$ in the diagram such that either $columnidx=columnidx',\\ constant\\le constant'$ or $columnidx\\le columnidx',\\ constant=constant'$ (including $roundness$ itself). Then the number of standard Young tableaux for this diagram equals\n\\[\n\\frac{voidsize!}{\\prod_{roundness} linegapp}.\n\\]\nFor a proof along the lines of the argument given in the previous remark, see: Kenneth Glass and Chi-Keung Ng, \"A simple proof of the hook length formula\", \\textit{American Mathematical Monthly} \\textbf{111} (2004), 700--704."
    },
    "garbled_string": {
      "map": {
        "S": "qzxwvtnp",
        "T": "hjgrksla",
        "j": "lmpqwert",
        "i": "vbnmhgfd",
        "a": "cxzasdwe",
        "b": "plmoknij",
        "c": "uytredfv",
        "d": "zxcvbnml",
        "s": "qwertyui",
        "f": "asdfghjk",
        "n": "poiulkjm",
        "n_1": "qazxswed",
        "n_2": "wsxedcrf",
        "n_3": "edcrfvtg",
        "N": "rfvtgbyh",
        "h_s": "tgbnhyuj"
      },
      "question": "Let $qzxwvtnp$ be the set of bijections\n\\[\nhjgrksla \\colon \\{1,2,3\\} \\times \\{1,2,\\dots,2024\\} \\to \\{1,2,\\dots,6072\\}\n\\]\nsuch that $hjgrksla(1,lmpqwert) < hjgrksla(2,lmpqwert) < hjgrksla(3,lmpqwert)$ for all $lmpqwert \\in \\{1,2,\\dots,2024\\}$ and $hjgrksla(vbnmhgfd,lmpqwert) < hjgrksla(vbnmhgfd,lmpqwert+1)$ for all $vbnmhgfd \\in \\{1,2,3\\}$ and $lmpqwert \\in \\{1,2,\\dots,2023\\}$. Do there exist $cxzasdwe$ and $uytredfv$ in \\{1,2,3\\} and $plmoknij$ and $zxcvbnml$ in \\{1,2,\\dots,2024\\} such that the fraction of elements $hjgrksla$ in $qzxwvtnp$ for which $hjgrksla(cxzasdwe,plmoknij) < hjgrksla(uytredfv,zxcvbnml)$ is at least $1/3$ and at most $2/3$?",
      "solution": "Yes, such $cxzasdwe,plmoknij,uytredfv,zxcvbnml$ exist: we take\n\\[\n(cxzasdwe,plmoknij) = (2,1), \\qquad (uytredfv,zxcvbnml) = (1,2).\n\\]\nWe will represent $hjgrksla$ as an $3 \\times poiulkjm$ array (3 rows, $poiulkjm$ columns) of integers  in which each of $1,\\dots,3poiulkjm$ occurs exactly once and the rows and columns are strictly increasing; we will specialize to $poiulkjm=2024$ at the end.\n\nWe first note that $hjgrksla(1,1) = 1$ and $2 \\in \\{hjgrksla(1,2), hjgrksla(2,1)\\}$. From this, it follows that $hjgrksla(2,1) < hjgrksla(1,2)$ if and only if $hjgrksla(2,1) = 2$.\n\nWe next recall a restricted form of the \\emph{hook length formula} (see the first remark for a short proof of this restricted version and the second remark for the statement of the general formula). Consider more generally an array consisting of (up to) three rows of lengths $qazxswed\\geq wsxedcrf \\geq edcrfvtg \\geq 0$, aligned at the left. Let $asdfghjk(qazxswed,wsxedcrf,edcrfvtg)$ be the number of ways to fill this array with a permutation of the numbers $1,\\dots,qazxswed+wsxedcrf+edcrfvtg$ in such a way that each row increases from left to right and each column increases from top to bottom. The hook length formula then shows that $asdfghjk(qazxswed,wsxedcrf,edcrfvtg)$ equals\n\\[\n\\frac{(qazxswed-wsxedcrf+1)(qazxswed-edcrfvtg+2)(wsxedcrf-edcrfvtg+1) (qazxswed+wsxedcrf+edcrfvtg)!}\n{(qazxswed+2)! (wsxedcrf+1)! edcrfvtg!}.\n\\]\n\nWe then note that if $hjgrksla(2,1) = 2$, we obtain a array with row lengths $poiulkjm, poiulkjm-1, poiulkjm-1$ by removing 1 and 2, relabeling each remaining $vbnmhgfd$ as $3poiulkjm+1-vbnmhgfd$, and reflecting in both axes. The probability that $hjgrksla(2,1) < hjgrksla(1,2)$ is thus\n\\begin{align*}\n\\frac{asdfghjk(poiulkjm,poiulkjm-1,poiulkjm-1)}{asdfghjk(poiulkjm,poiulkjm,poiulkjm)}\n&= \n\\frac{(2)(3)(poiulkjm+1)poiulkjm}{(1)(2) (3poiulkjm)(3poiulkjm-1)} \\\\\n&= \\frac{poiulkjm+1}{3poiulkjm-1} = \\frac{1}{3} + \\frac{4}{9poiulkjm-3};\n\\end{align*}\nthis is always greater than $\\frac{1}{3}$, and for $poiulkjm =2024$ it is visibly less than $\\frac{2}{3}$.\n\n\\noindent\n\\textbf{Remark.}\nWe prove the claimed formula for $asdfghjk(qazxswed,wsxedcrf,edcrfvtg)$ by induction on $qazxswed+wsxedcrf+edcrfvtg$. To begin with, if $wsxedcrf = edcrfvtg = 0$, then the desired count is indeed $asdfghjk(qazxswed, 0, 0) = 1$. Next, suppose $wsxedcrf > 0, edcrfvtg = 0$. The entry $qazxswed + wsxedcrf$ must go at the end of either the first or second row; counting ways to complete the diagram from these starting points yields\n\\[\nasdfghjk(qazxswed,wsxedcrf,0) = asdfghjk(qazxswed-1,wsxedcrf,0) + asdfghjk(qazxswed,wsxedcrf-1,0).\n\\]\n(This works even if $qazxswed = wsxedcrf$, in which case the first row is not an option but correspondingly $asdfghjk(wsxedcrf-1,wsxedcrf,0) = 0$.) The induction step then follows from the identity\n\\[\n\\frac{(qazxswed-wsxedcrf)(qazxswed+1) + (qazxswed-wsxedcrf+2)wsxedcrf}{(qazxswed-wsxedcrf+1)(qazxswed+wsxedcrf)}  = 1.\n\\]\n(As an aside, the case $qazxswed = wsxedcrf, edcrfvtg = 0$ recovers a standard interpretation of the Catalan numbers.)\n\nFinally, suppose $edcrfvtg > 0$. We then have\n\\begin{align*}\n&asdfghjk(qazxswed,wsxedcrf,edcrfvtg) \\\\\n&= \nasdfghjk(qazxswed-1,wsxedcrf,edcrfvtg) + asdfghjk(qazxswed,wsxedcrf-1,edcrfvtg) + asdfghjk(qazxswed,wsxedcrf,edcrfvtg-1),\n\\end{align*}\nand the induction step now reduces to the algebraic identity\n\\begin{align*}\n&\\frac{(qazxswed-wsxedcrf)(qazxswed-edcrfvtg+1)(qazxswed+2)}{(qazxswed-wsxedcrf+1)(qazxswed-edcrfvtg+2)(qazxswed+wsxedcrf+edcrfvtg)} \\\\\n&+ \\frac{(qazxswed-wsxedcrf+2)(wsxedcrf-edcrfvtg)(wsxedcrf+1)}{(qazxswed-wsxedcrf+1)(wsxedcrf-edcrfvtg+1)(qazxswed+wsxedcrf+edcrfvtg)} \\\\\n&+ \\frac{(qazxswed-edcrfvtg+3)(wsxedcrf-edcrfvtg+2)edcrfvtg}{(qazxswed-edcrfvtg+2)(wsxedcrf-edcrfvtg+1)(qazxswed+wsxedcrf+edcrfvtg)}\n= 1.\n\\end{align*}\n\n\\noindent\n\\textbf{Remark.}\nWe formulate the general hook length formula in standard terminology. Let $rfvtgbyh$ be a positive integer, and consider a semi-infinite checkerboard with top and left edges. A \\emph{Ferrers diagram} is a finite subset of the squares of the board which is closed under taking a unit step towards either edge. Given a Ferrers diagram with $rfvtgbyh$ squares, a \\emph{standard Young tableau} for this diagram is a bijection of the squares of the diagram with the integers $1,\\dots,rfvtgbyh$ such that the numbers always increase under taking a unit step away from either edge. \n\n\nFor each square $qwertyui = (vbnmhgfd,lmpqwert)$ in the diagram, the \\emph{hook length} $tgbnhyuj$ of $qwertyui$ is the number of squares $(vbnmhgfd',lmpqwert')$ in the diagram such that either $vbnmhgfd=vbnmhgfd', lmpqwert\\leq lmpqwert'$ or $vbnmhgfd\\leq vbnmhgfd',lmpqwert=lmpqwert'$ (including $qwertyui$ itself). Then the number of standard Young tableaux for this diagram equals\n\\[\n\\frac{rfvtgbyh!}{\\prod_{qwertyui} tgbnhyuj}.\n\\]\nFor a proof along the lines of the argument given in the previous remark, see: Kenneth Glass and Chi-Keung Ng, A simple proof of the hook length formula, \\textit{American Mathematical Monthly} \\textbf{111} (2004), 700--704."
    },
    "kernel_variant": {
      "question": "Let $S$ be the set of bijections\\[\nT:\n\\{1,2,3\\}\\times\\{1,2,\\dots ,1729\\}\\longrightarrow\\{1,2,\\dots ,5187\\}\n\\]having the two monotonicity properties\n\\[\nT(1,j)<T(2,j)<T(3,j)\\qquad(1\\le j\\le 1729),\\qquad\\qquad\nT(i,j)<T(i,j+1)\\qquad(1\\le i\\le3,\\,1\\le j\\le1728).\n\\]\nShow that there exist $a,c\\in\\{1,2,3\\}$ and $b,d\\in\\{1,2,\\dots ,1729\\}$ such that the proportion of tableaux $T\\in S$ satisfying $T(a,b)<T(c,d)$ is at least $\\dfrac13$ and at most $\\dfrac23$.\n",
      "solution": "Fix N=1729, and view each T∈S as a 3×N array of the numbers 1,…,3N=5187, strictly increasing in each row and each column.  We claim that choosing (a,b)=(2,1) and (c,d)=(1,2) works.\n\n1.  Because the smallest entry must occupy (1,1), we have T(1,1)=1, and hence the entry 2 lies either at (2,1) or at (1,2).  Clearly\n   T(2,1)<T(1,2)   ⇔   T(2,1)=2.\n\n2.  Let f(n_1,n_2,n_3) be the number of ways to fill a left‐justified 3‐row shape of row‐lengths n_1≥n_2≥n_3 with the numbers 1,…,n_1+n_2+n_3 so that rows increase left‐to‐right and columns increase top‐to‐bottom.  By a standard hook‐length argument,\n   f(n_1,n_2,n_3)\n   =\n   \\frac{(n_1−n_2+1)(n_1−n_3+2)(n_2−n_3+1)\times(n_1+n_2+n_3)!}{(n_1+2)!\bigl(n_2+1\bigr)!\bigl(n_3\bigr)!}.\n\n3.  Removing the entries 1 and 2 from T when T(2,1)=2 produces a shape of row‐lengths (N,N−1,N−1).  Thus\n   #{T∈S : T(2,1)=2}\n   =f(N,N−1,N−1),\n   #S=f(N,N,N).\n\n4.  A straightforward cancellation shows\n   f(N,N−1,N−1)/f(N,N,N) = (N+1)/(3N−1).\n\n5.  Substituting N=1729 gives\n   (1729+1)/(3·1729−1) = 1730/5186 = 865/2593,\nwhich indeed satisfies\n   1/3 < 865/2593 < 2/3.\n\nTherefore for (a,b)=(2,1) and (c,d)=(1,2), the proportion of T∈S with T(a,b)<T(c,d) lies between 1/3 and 2/3, as required.",
      "_meta": {
        "core_steps": [
          "Note 1 must be at (1,1); strict monotonicity forces 2 to be either (2,1) or (1,2), so T(2,1)<T(1,2) ⇔ T(2,1)=2.",
          "Rephrase the probability as the ratio of tableaux in which 2 is at (2,1); after deleting 1,2 this is shape (n,n-1,n-1) versus all tableaux of shape (n,n,n).",
          "Count these shapes with the three-row hook-length formula.",
          "Compute the ratio (n+1)/(3n-1) and confirm it is in (1/3,2/3) for every n≥2 (hence for the given n)."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "number of columns (second index bound) in the array, denoted n",
            "original": "2024"
          },
          "slot2": {
            "description": "largest label = total entries = 3·n appearing in the tableau",
            "original": "6072"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}