{
  "index": "1956-A-5",
  "type": "COMB",
  "tag": [
    "COMB",
    "ALG"
  ],
  "difficulty": "",
  "question": "5. Given \\( n \\) objects arranged in a row. A subset of these objects is called unfriendly if no two of its elements are consecutive. Show that the number of unfriendly subsets each having \\( k \\) elements is\n\\[\n\\binom{n-k+1}{k}\n\\]",
  "solution": "Solution. For each subset \\( S \\) of an ordered set of \\( n \\) objects we form a linear arrangement of \\( A \\) 's and \\( B \\) 's by writing an \\( A \\) in positions corresponding to members of \\( S \\) and \\( B \\) 's elsewhere. We describe such an arrangement as unfriendly if no two \\( A \\) 's are consecutive. Then an unfriendly subset corresponds to an unfriendly arrangement, and we must show that the number of unfriendly arrangements of \\( k A \\) 's and \\( n-k B \\) 's is \\( \\binom{n-k+1}{k} \\).\n\nWe shall establish a bijective correspondence between all unfriendly arrangements of \\( k \\) 's and \\( n-k B \\) 's and all arrangements of \\( k A \\) 's and \\( n-2 k+1 B ' s \\).\n\nGiven an unfriendly arrangement of \\( k A \\) 's and \\( n-k B \\) 's, remove the \\( B \\) standing immediately to the right of each of the first \\( k-1 A \\) 's (i.e., all but the last \\( A \\) ). We obtain in this way an arrangement of \\( k A \\) 's and \\( n-2 k \\) \\( +1 B \\) 's.\n\nConversely, given an arrangement of \\( k A \\) 's and \\( n-2 k+1 B \\) 's, insert a \\( B \\) after each of the \\( A \\) 's but the last. We obtain in this way an unfriendly arrangement of \\( k A \\) 's and \\( n-k B \\) 's.\n\nObviously these two transformations are inverses of one another. Therefore the number of unfriendly arrangements of \\( k A \\) 's and \\( n-k B \\) 's is the same as the number of arrangements of \\( k A \\) 's and \\( n-2 k+1 B \\) 's, namely\n\\[\n\\binom{n-k+1}{k}\n\\]\n\nThis problem was treated by Irving Kaplansky, \"Solution of the problème des ménages, \"Bulletin of the American Mathematical Society, vol. 49 (1943), pages 784-785. The problem has been generalized by H. D. Abramson, \"On Selecting Separated Objects from a Row,\" American Mathematical Monthly, vol. 76 (1969), pages 1130-1131. The generalization requires greater separation between the elements in the unfriendly subset.",
  "vars": [
    "n",
    "k",
    "S",
    "A",
    "B"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "objectcount",
        "k": "subsetsize",
        "S": "selectedset",
        "A": "chosenmark",
        "B": "gapmark"
      },
      "question": "5. Given \\( objectcount \\) objects arranged in a row. A subset of these objects is called unfriendly if no two of its elements are consecutive. Show that the number of unfriendly subsets each having \\( subsetsize \\) elements is\n\\[\n\\binom{objectcount-subsetsize+1}{subsetsize}\n\\]",
      "solution": "Solution. For each subset \\( selectedset \\) of an ordered set of \\( objectcount \\) objects we form a linear arrangement of \\( chosenmark \\) 's and \\( gapmark \\) 's by writing a \\( chosenmark \\) in positions corresponding to members of \\( selectedset \\) and \\( gapmark \\) 's elsewhere. We describe such an arrangement as unfriendly if no two \\( chosenmark \\) 's are consecutive. Then an unfriendly subset corresponds to an unfriendly arrangement, and we must show that the number of unfriendly arrangements of \\( subsetsize chosenmark \\) 's and \\( objectcount-subsetsize gapmark \\) 's is \\( \\binom{objectcount-subsetsize+1}{subsetsize} \\).\n\nWe shall establish a bijective correspondence between all unfriendly arrangements of \\( subsetsize chosenmark \\) 's and \\( objectcount-subsetsize gapmark \\) 's and all arrangements of \\( subsetsize chosenmark \\) 's and \\( objectcount-2 subsetsize+1 gapmark ' s \\).\n\nGiven an unfriendly arrangement of \\( subsetsize chosenmark \\) 's and \\( objectcount-subsetsize gapmark \\) 's, remove the \\( gapmark \\) standing immediately to the right of each of the first \\( subsetsize-1 chosenmark \\) 's (i.e., all but the last \\( chosenmark \\) ). We obtain in this way an arrangement of \\( subsetsize chosenmark \\) 's and \\( objectcount-2 subsetsize +1 gapmark \\) 's.\n\nConversely, given an arrangement of \\( subsetsize chosenmark \\) 's and \\( objectcount-2 subsetsize+1 gapmark \\) 's, insert a \\( gapmark \\) after each of the \\( chosenmark \\) 's but the last. We obtain in this way an unfriendly arrangement of \\( subsetsize chosenmark \\) 's and \\( objectcount-subsetsize gapmark \\) 's.\n\nObviously these two transformations are inverses of one another. Therefore the number of unfriendly arrangements of \\( subsetsize chosenmark \\) 's and \\( objectcount-subsetsize gapmark \\) 's is the same as the number of arrangements of \\( subsetsize chosenmark \\) 's and \\( objectcount-2 subsetsize+1 gapmark \\) 's, namely\n\\[\n\\binom{objectcount-subsetsize+1}{subsetsize}\n\\]\n\nThis problem was treated by Irving Kaplansky, \"Solution of the probl\u0000e8me des m\u0000e9nages,\" Bulletin of the American Mathematical Society, vol. 49 (1943), pages 784-785. The problem has been generalized by H. D. Abramson, \"On Selecting Separated Objects from a Row,\" American Mathematical Monthly, vol. 76 (1969), pages 1130-1131. The generalization requires greater separation between the elements in the unfriendly subset."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "pineapple",
        "k": "lighthouse",
        "S": "corridor",
        "A": "meteorite",
        "B": "sailboat"
      },
      "question": "5. Given \\( pineapple \\) objects arranged in a row. A subset of these objects is called unfriendly if no two of its elements are consecutive. Show that the number of unfriendly subsets each having \\( lighthouse \\) elements is\n\\[\n\\binom{pineapple-lighthouse+1}{lighthouse}\n\\]",
      "solution": "Solution. For each subset \\( corridor \\) of an ordered set of \\( pineapple \\) objects we form a linear arrangement of \\( meteorite \\) 's and \\( sailboat \\) 's by writing an \\( meteorite \\) in positions corresponding to members of \\( corridor \\) and \\( sailboat \\) 's elsewhere. We describe such an arrangement as unfriendly if no two \\( meteorite \\) 's are consecutive. Then an unfriendly subset corresponds to an unfriendly arrangement, and we must show that the number of unfriendly arrangements of \\( lighthouse\\, meteorite \\) 's and \\( pineapple-lighthouse\\, sailboat \\) 's is \\( \\binom{pineapple-lighthouse+1}{lighthouse} \\).\n\nWe shall establish a bijective correspondence between all unfriendly arrangements of \\( lighthouse \\) 's and \\( pineapple-lighthouse\\, sailboat \\) 's and all arrangements of \\( lighthouse\\, meteorite \\) 's and \\( pineapple-2 lighthouse+1\\, sailboat ' s \\).\n\nGiven an unfriendly arrangement of \\( lighthouse\\, meteorite \\) 's and \\( pineapple-lighthouse\\, sailboat \\) 's, remove the \\( sailboat \\) standing immediately to the right of each of the first \\( lighthouse-1\\, meteorite \\) 's (i.e., all but the last \\( meteorite \\) ). We obtain in this way an arrangement of \\( lighthouse\\, meteorite \\) 's and \\( pineapple-2 lighthouse \\) \\( +1\\, sailboat \\) 's.\n\nConversely, given an arrangement of \\( lighthouse\\, meteorite \\) 's and \\( pineapple-2 lighthouse+1\\, sailboat \\) 's, insert a \\( sailboat \\) after each of the \\( meteorite \\) 's but the last. We obtain in this way an unfriendly arrangement of \\( lighthouse\\, meteorite \\) 's and \\( pineapple-lighthouse\\, sailboat \\) 's.\n\nObviously these two transformations are inverses of one another. Therefore the number of unfriendly arrangements of \\( lighthouse\\, meteorite \\) 's and \\( pineapple-lighthouse\\, sailboat \\) 's is the same as the number of arrangements of \\( lighthouse\\, meteorite \\) 's and \\( pineapple-2 lighthouse+1\\, sailboat \\) 's, namely\n\\[\n\\binom{pineapple-lighthouse+1}{lighthouse}\n\\]\n\nThis problem was treated by Irving Kaplansky, \"Solution of the problème des ménages, \"Bulletin of the American Mathematical Society, vol. 49 (1943), pages 784-785. The problem has been generalized by H. D. Abramson, \"On Selecting Separated Objects from a Row,\" American Mathematical Monthly, vol. 76 (1969), pages 1130-1131. The generalization requires greater separation between the elements in the unfriendly subset."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "emptiness",
        "k": "wholeness",
        "S": "superset",
        "A": "excluded",
        "B": "included"
      },
      "question": "5. Given \\( \\emptiness \\) objects arranged in a row. A subset of these objects is called unfriendly if no two of its elements are consecutive. Show that the number of unfriendly subsets each having \\( \\wholeness \\) elements is\n\\[\n\\binom{\\emptiness-\\wholeness+1}{\\wholeness}\n\\]",
      "solution": "Solution. For each subset \\( \\superset \\) of an ordered set of \\( \\emptiness \\) objects we form a linear arrangement of \\( \\excluded \\)'s and \\( \\included \\)'s by writing an \\( \\excluded \\) in positions corresponding to members of \\( \\superset \\) and \\( \\included \\)'s elsewhere. We describe such an arrangement as unfriendly if no two \\( \\excluded \\)'s are consecutive. Then an unfriendly subset corresponds to an unfriendly arrangement, and we must show that the number of unfriendly arrangements of \\( \\wholeness \\excluded \\)'s and \\( \\emptiness-\\wholeness \\included \\)'s is \\( \\binom{\\emptiness-\\wholeness+1}{\\wholeness} \\).\n\nWe shall establish a bijective correspondence between all unfriendly arrangements of \\( \\wholeness \\excluded \\)'s and \\( \\emptiness-\\wholeness \\included \\)'s and all arrangements of \\( \\wholeness \\excluded \\)'s and \\( \\emptiness-2\\wholeness+1 \\included \\)'s.\n\nGiven an unfriendly arrangement of \\( \\wholeness \\excluded \\)'s and \\( \\emptiness-\\wholeness \\included \\)'s, remove the \\( \\included \\) standing immediately to the right of each of the first \\( \\wholeness-1 \\excluded \\)'s (i.e., all but the last \\( \\excluded \\)). We obtain in this way an arrangement of \\( \\wholeness \\excluded \\)'s and \\( \\emptiness-2\\wholeness+1 \\included \\)'s.\n\nConversely, given an arrangement of \\( \\wholeness \\excluded \\)'s and \\( \\emptiness-2\\wholeness+1 \\included \\)'s, insert a \\( \\included \\) after each of the \\( \\excluded \\)'s but the last. We obtain in this way an unfriendly arrangement of \\( \\wholeness \\excluded \\)'s and \\( \\emptiness-\\wholeness \\included \\)'s.\n\nObviously these two transformations are inverses of one another. Therefore the number of unfriendly arrangements of \\( \\wholeness \\excluded \\)'s and \\( \\emptiness-\\wholeness \\included \\)'s is the same as the number of arrangements of \\( \\wholeness \\excluded \\)'s and \\( \\emptiness-2\\wholeness+1 \\included \\)'s, namely\n\\[\n\\binom{\\emptiness-\\wholeness+1}{\\wholeness}\n\\]"
    },
    "garbled_string": {
      "map": {
        "n": "vhnoegrt",
        "k": "czmpyldf",
        "S": "qikzruva",
        "A": "xobkfwey",
        "B": "jtrencuo"
      },
      "question": "5. Given \\( vhnoegrt \\) objects arranged in a row. A subset of these objects is called unfriendly if no two of its elements are consecutive. Show that the number of unfriendly subsets each having \\( czmpyldf \\) elements is\n\\[\n\\binom{vhnoegrt-czmpyldf+1}{czmpyldf}\n\\]\n",
      "solution": "Solution. For each subset \\( qikzruva \\) of an ordered set of \\( vhnoegrt \\) objects we form a linear arrangement of \\( xobkfwey \\) 's and \\( jtrencuo \\) 's by writing an \\( xobkfwey \\) in positions corresponding to members of \\( qikzruva \\) and \\( jtrencuo \\) 's elsewhere. We describe such an arrangement as unfriendly if no two \\( xobkfwey \\) 's are consecutive. Then an unfriendly subset corresponds to an unfriendly arrangement, and we must show that the number of unfriendly arrangements of \\( czmpyldf xobkfwey \\) 's and \\( vhnoegrt-czmpyldf jtrencuo \\) 's is \\( \\binom{vhnoegrt-czmpyldf+1}{czmpyldf} \\).\n\nWe shall establish a bijective correspondence between all unfriendly arrangements of \\( czmpyldf \\) 's and \\( vhnoegrt-czmpyldf jtrencuo \\) 's and all arrangements of \\( czmpyldf xobkfwey \\) 's and \\( vhnoegrt-2 czmpyldf+1 jtrencuo ' s \\).\n\nGiven an unfriendly arrangement of \\( czmpyldf xobkfwey \\) 's and \\( vhnoegrt-czmpyldf jtrencuo \\) 's, remove the \\( jtrencuo \\) standing immediately to the right of each of the first \\( czmpyldf-1 xobkfwey \\) 's (i.e., all but the last \\( xobkfwey \\) ). We obtain in this way an arrangement of \\( czmpyldf xobkfwey \\) 's and \\( vhnoegrt-2 czmpyldf \\)  \\( +1 jtrencuo \\) 's.\n\nConversely, given an arrangement of \\( czmpyldf xobkfwey \\) 's and \\( vhnoegrt-2 czmpyldf+1 jtrencuo \\) 's, insert a \\( jtrencuo \\) after each of the \\( xobkfwey \\) 's but the last. We obtain in this way an unfriendly arrangement of \\( czmpyldf xobkfwey \\) 's and \\( vhnoegrt-czmpyldf jtrencuo \\) 's.\n\nObviously these two transformations are inverses of one another. Therefore the number of unfriendly arrangements of \\( czmpyldf xobkfwey \\) 's and \\( vhnoegrt-czmpyldf jtrencuo \\) 's is the same as the number of arrangements of \\( czmpyldf xobkfwey \\) 's and \\( vhnoegrt-2 czmpyldf+1 jtrencuo \\) 's, namely\n\\[\n\\binom{vhnoegrt-czmpyldf+1}{czmpyldf}\n\\]\n\nThis problem was treated by Irving Kaplansky, \"Solution of the problème des ménages,\" Bulletin of the American Mathematical Society, vol. 49 (1943), pages 784-785. The problem has been generalized by H. D. Abramson, \"On Selecting Separated Objects from a Row,\" American Mathematical Monthly, vol. 76 (1969), pages 1130-1131. The generalization requires greater separation between the elements in the unfriendly subset."
    },
    "kernel_variant": {
      "question": "Let  \n\n$\\bullet\\; n\\ge 1$  (length of the row),  \n\n$\\bullet\\; s\\ge 2$  (required minimal separation),  \n\n$\\bullet\\; k\\ge 0$  (prescribed size of the subset),  \n\n$\\bullet\\; q\\ge 2$  (a modulus), and  \n\n$\\bullet\\; r$ with $0\\le r<q$  \n\nbe fixed integers.  \n\nA subset $S\\subseteq\\{1,2,\\dots ,n\\}$ is called $(s,q,r)$–aloof if  \n\n1. $\\lvert S\\rvert =k$,  \n\n2. $\\lvert i-j\\rvert \\ge s$ for every two distinct $i,j\\in S$, and  \n\n3. $\\displaystyle\\sum_{i\\in S} i\\equiv r\\pmod q$.  \n\nFor such data set  \n\n\\[\nA_{n,k}^{(s,q,r)}=\\#\\bigl\\{S\\subseteq \\{1,\\dots ,n\\}\\;:\\;S\\text{ is }(s,q,r)\\text{–aloof}\\bigr\\}.\n\\]\n\n(a) Prove the master identity  \n\n\\[\nA_{n,k}^{(s,q,r)}\n=\\frac1q\\sum_{t=0}^{q-1}\\;\n\\omega^{-tr}\\,\n\\omega^{t\\,(s-1)k(k-1)/2}\\,\n\\omega^{t\\,k(k+1)/2}\\,\n\\bigl[{M\\choose k}\\bigr]_{\\omega^{t}},\\tag{$\\star$}\n\\]\n\nwhere  \n\n\\[\nM:=n-(s-1)(k-1),\\qquad \n\\omega:=e^{2\\pi i/q},\\qquad\n\\bigl[{M\\choose k}\\bigr]_{x}:=\n\\frac{(1-x^{M})(1-x^{M-1})\\cdots(1-x^{M-k+1})}\n     {(1-x)(1-x^{2})\\cdots(1-x^{k})}.\n\\]\n\n(Throughout, the right–hand side is interpreted as $0$ if $M<k$.)\n\n(b) Derive from $(\\star)$ explicit closed formulas in the two distinguished situations  \n\n(i) $q=2$ (parity restriction);  \n\n(ii) $s=2$ \\emph{with $q$ prime} (classical “no two chosen positions are consecutive”, enriched by a prime–modulus congruence).",
      "solution": "Throughout set  \n\n\\[\nM:=n-(s-1)(k-1).\\tag{1}\n\\]\n\nWhen $M<k$ both sides of $(\\star)$ vanish, so we henceforth assume $M\\ge k$.\n\n\\textbf{Part (a).  Proof of $(\\star)$.}  \n\nThe argument proceeds through six standard steps.\n\n1. \\emph{Compression of the spacing condition.}  \nFor $S=\\{x_{1}<\\dots <x_{k}\\}$ put  \n\n\\[\ny_i:=x_i-(i-1)(s-1)\\qquad(1\\le i\\le k).\\tag{2}\n\\]\n\nSince $x_{i+1}-x_i\\ge s$, the integers $y_1<\\dots <y_k$ form an ordinary $k$–subset of $\\{1,\\dots ,M\\}$, and \\eqref{2} is a bijection.\n\n2. \\emph{Translation of the congruence.}  \nSumming \\eqref{2} gives  \n\n\\[\n\\sum_{i=1}^{k}x_i=\\sum_{i=1}^{k}y_i+(s-1)\\frac{k(k-1)}{2},\\tag{3}\n\\]\nso  \n\n\\[\n\\sum_{i=1}^{k}y_i\\equiv r-\\frac{(s-1)k(k-1)}{2}\\pmod q.\\tag{4}\n\\]\nWrite  \n\n\\[\nc\\equiv r-\\frac{(s-1)k(k-1)}{2}\\pmod q.\\tag{5}\n\\]\n\nWe thus count $k$–subsets of $\\{1,\\dots ,M\\}$ whose element–sum is congruent to $c$ modulo $q$.\n\n3. \\emph{A bivariate generating series.}  \nIntroduce indeterminates $u$ (subset size) and $z$ (element sum) and define  \n\n\\[\nF(u,z):=\\prod_{j=1}^{M}\\bigl(1+u\\,z^{\\,j}\\bigr).\\tag{6}\n\\]\n\nThe coefficient of $u^{k}z^{m}$ in $F$ counts the desired subsets with sum $m$.\n\n4. \\emph{Root–of–unity filter.}  \nWith $\\omega=e^{2\\pi i/q}$ we project onto $m\\equiv c\\pmod q$:\n\n\\[\nA_{n,k}^{(s,q,r)}\n=\\frac1q\\sum_{t=0}^{q-1}\\omega^{-tc}\\,\\bigl[u^{k}\\bigr]F\\!\\bigl(u,z=\\omega^{t}\\bigr).\\tag{7}\n\\]\n\n5. \\emph{Extraction of the $u^{k}$–coefficient.}  \nFix $t$ and abbreviate $\\xi:=\\omega^{t}$, $v:=u\\xi$.  Gauss’ $q$–binomial theorem yields  \n\n\\[\n\\prod_{j=0}^{M-1}(1+v\\,\\xi^{\\,j})\n     =\\sum_{\\ell=0}^{M}v^{\\ell}\\xi^{\\,\\ell(\\ell-1)/2}\n       \\bigl[{M\\choose \\ell}\\bigr]_{\\xi}.\\tag{8}\n\\]\n\nReplacing $v$ by $u\\xi$ and extracting $u^{k}$ we get  \n\n\\[\n\\bigl[u^{k}\\bigr]F(u,\\xi)=\\xi^{\\,k(k+1)/2}\\,\n                           \\bigl[{M\\choose k}\\bigr]_{\\xi}.\\tag{9}\n\\]\n\n6. \\emph{Assembly.}  Insert \\eqref{9} into \\eqref{7} and recall \\eqref{5}:\n\n\\[\n\\begin{aligned}\nA_{n,k}^{(s,q,r)}\n&=\\frac1q\\sum_{t=0}^{q-1}\n   \\omega^{-tc}\\,\n   \\omega^{\\,t\\,k(k+1)/2}\\,\n   \\bigl[{M\\choose k}\\bigr]_{\\omega^{t}} \\\\[4pt]\n&=\\frac1q\\sum_{t=0}^{q-1}\n   \\omega^{-tr}\\,\n   \\omega^{\\,t\\,(s-1)k(k-1)/2}\\,\n   \\omega^{\\,t\\,k(k+1)/2}\\,\n   \\bigl[{M\\choose k}\\bigr]_{\\omega^{t}},\n\\end{aligned}\n\\]\n\nwhich is precisely $(\\star)$.  Part (a) is complete.\n\n\\bigskip\n\\textbf{Part (b).  Specialisations.}\n\n\\smallskip\n\\textbf{(i)  Modulus $q=2$.}  \nNow $\\omega=-1$ and $t\\in\\{0,1\\}$.  Put  \n\n\\[\nE:=-r+\\frac{(s-1)k(k-1)}{2}+\\frac{k(k+1)}{2}.\\tag{10}\n\\]\n\nFormula $(\\star)$ gives  \n\n\\[\nA_{n,k}^{(s,2,r)}\n=\\frac12\\,\\binom{M}{k}\n+\\frac12\\,(-1)^{E}\\,\n          \\bigl[{M\\choose k}\\bigr]_{-1}.\\tag{11}\n\\]\n\nThe classical evaluation  \n\n\\[\n\\bigl[{M\\choose k}\\bigr]_{-1}\n=\\begin{cases}\n0, & k(M-k)\\text{ odd},\\\\[8pt]\n\\displaystyle\\binom{\\lfloor M/2\\rfloor}{\\lfloor k/2\\rfloor}, & k(M-k)\\text{ even},\n\\end{cases}\\tag{12}\n\\]\n\nturns \\eqref{11} into a fully explicit closed form.\n\n\\smallskip\n\\textbf{(ii)  Separation $s=2$ and \\emph{prime} modulus $q$.}  \n\nHere $M=n-k+1=:N$, so $(\\star)$ reads  \n\n\\[\nA_{n,k}^{(2,q,r)}\n=\\frac1q\\sum_{t=0}^{q-1}\n  \\omega^{\\,t\\,(k^{2}-r)}\\,\n  \\bigl[{N\\choose k}\\bigr]_{\\omega^{t}}.\\tag{13}\n\\]\n\n\\emph{Separating the contribution $t=0$.}  Since $\\omega^{0}=1$,  \n\n\\[\n\\bigl[{N\\choose k}\\bigr]_{1}=\\binom{N}{k},\n\\]\nwhence  \n\n\\[\nA_{n,k}^{(2,q,r)}\n=\\frac{1}{q}\\binom{N}{k}\n+\\frac{1}{q}\\sum_{t=1}^{q-1}\n  \\omega^{\\,t\\,(k^{2}-r)}\\,\n  \\bigl[{N\\choose k}\\bigr]_{\\omega^{t}}.\\tag{14}\n\\]\n\n\\emph{$q$–Lucas decomposition for $t\\neq0$.}  \nBecause $q$ is prime, every $\\omega^{t}$ ($t=1,\\dots ,q-1$) is a primitive $q$–th root; hence $\\omega^{t\\,j}\\neq1$ for $1\\le j\\le q-1$.  Write  \n\n\\[\nN=qM_{1}+M_{0},\\qquad \nk=qk_{1}+k_{0},\\qquad 0\\le M_{0},k_{0}<q.\\tag{15}\n\\]\n\n$q$–Lucas’ theorem gives  \n\n\\[\n\\bigl[{N\\choose k}\\bigr]_{\\omega^{t}}\n=\\binom{M_{1}}{k_{1}}\\,\n \\bigl[{M_{0}\\choose k_{0}}\\bigr]_{\\omega^{t}}.\\tag{16}\n\\]\n\n\\emph{Evaluating the small Gaussian coefficient.}  \nBecause $M_{0},k_{0}<q$ and $\\omega^{t\\,j}\\neq1$ for $1\\le j\\le q-1$,  \n\n\\[\n\\bigl[{M_{0}\\choose k_{0}}\\bigr]_{\\omega^{t}}\n=\\begin{cases}\n0, & k_{0}>M_{0},\\\\[10pt]\n\\displaystyle\\prod_{j=1}^{k_{0}}\n     \\frac{1-\\omega^{t\\,(M_{0}-k_{0}+j)}}{1-\\omega^{t\\,j}},\n & k_{0}\\le M_{0}.\n\\end{cases}\\tag{17}\n\\]\n\nAll denominators are non–zero, so \\eqref{17} is well defined.\n\n\\emph{Putting everything together.}  Substituting \\eqref{16}–\\eqref{17} into \\eqref{14} yields  \n\n\\[\n\\boxed{\\;\nA_{n,k}^{(2,q,r)}=\n\\frac{1}{q}\\binom{n-k+1}{k}\\;+\\;\n\\frac{\\binom{M_{1}}{k_{1}}}{q}\\,\n\\mathbf 1_{\\,k_{0}\\le M_{0}}\\,\n\\sum_{t=1}^{q-1}\\omega^{\\,t\\,(k^{2}-r)}\n\\prod_{j=1}^{k_{0}}\n     \\frac{1-\\omega^{t\\,(M_{0}-k_{0}+j)}}{1-\\omega^{t\\,j}}\n}\\tag{18}\n\\]\n\nwhere $\\mathbf 1_{\\,\\ast}$ is the indicator ($1$ if the condition holds, $0$ otherwise).  \nFormula \\eqref{18} expresses $A_{n,k}^{(2,q,r)}$ as a finite sum of $(q-1)$ explicit products, each involving at most $2k_{0}$ ordinary complex numbers, and is therefore a bona–fide closed form.\n\n\\smallskip\n\\textbf{Two immediate corollaries.}\n\n1.  \\emph{The case $k\\equiv 0\\pmod q$ (hence $k_{0}=0$).}  \nNow the empty product equals $1$, and\n\n\\[\n\\sum_{t=1}^{q-1}\\omega^{\\,t\\,(k^{2}-r)}\n=\n\\begin{cases}\nq-1,& k^{2}\\equiv r\\pmod q,\\\\\n-1,& k^{2}\\not\\equiv r\\pmod q.\n\\end{cases}\n\\]\n\nHence \\eqref{18} gives  \n\n\\[\nA_{n,k}^{(2,q,r)}\n=\n\\begin{cases}\n\\displaystyle\n\\binom{M_{1}}{k_{1}}\n+\\dfrac1q\\Bigl(\\binom{N}{k}-\\binom{M_{1}}{k_{1}}\\Bigr), & k^{2}\\equiv r\\pmod q,\\\\[10pt]\n\\displaystyle\n\\dfrac1q\\Bigl(\\binom{N}{k}-\\binom{M_{1}}{k_{1}}\\Bigr), & k^{2}\\not\\equiv r\\pmod q.\n\\end{cases}\\tag{19}\n\\]\n\nThis corrects the erroneous omission of the additional term $\\dfrac1q\\bigl(\\binom{N}{k}-\\binom{M_{1}}{k_{1}}\\bigr)$ in the original draft.\n\n2.  \\emph{Compatibility with $q=2$.}  \nSpecialising \\eqref{18} to $q=2$ gives precisely \\eqref{11}–\\eqref{12}, so formula \\eqref{18} is consistent with part (i).\n\n\\bigskip\nThe argument is complete.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.480125",
        "was_fixed": false,
        "difficulty_analysis": "• Additional parameter s ≥ 2 forces a minimum spacing larger than 1, so the simple “delete the k−1 B’s” trick of the original solution no longer suffices; one must perform the more involved compression (1).  \n• An arithmetic congruence on the sum of the chosen indices introduces number-theoretic conditions absent from the original task.  \n• Solving the problem now needs three advanced tools:  \n 1. A stars-and-bars–type compression for s-spacing;  \n 2. q–analogue combinatorics (Gaussian binomials) to encode sums;  \n 3. A discrete Fourier / root–of–unity filter to isolate a specified residue class.  \n• The final closed form (★) intertwines graph-theoretic independence, additive number theory, and algebraic combinatorics; it cannot be obtained by elementary pattern matching or direct counting.  \n• Even the two “simpler” corollaries demand familiarity with parity arguments or with evaluating Gaussian coefficients at q = −1, again well beyond the original binomial-coefficient level.\n\nHence the enhanced kernel variant is substantially more technical and conceptually deeper than both the original problem and its current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Fix five integers  \n\n$\\bullet\\; n\\ge 1$  (length of the row),  \n$\\bullet\\; s\\ge 2$  (required minimal separation),  \n$\\bullet\\; k\\ge 0$  (prescribed size of the subset),  \n$\\bullet\\; q\\ge 2$  (a modulus), and  \n$\\bullet\\; r$ with $0\\le r<q$.  \n\nA subset $S\\subseteq\\{1,2,\\dots ,n\\}$ is called $(s,q,r)$–aloof if  \n\n1. $\\lvert S\\rvert =k$,  \n2. $\\lvert i-j\\rvert \\ge s$ for every two distinct $i,j\\in S$, and  \n3. $\\displaystyle\\sum_{i\\in S} i\\equiv r\\pmod q$.  \n\nWrite  \n\n\\[\nA_{n,k}^{(s,q,r)}=\\#\\bigl\\{S\\subseteq \\{1,\\dots ,n\\}\\;:\\;S\\text{ is }(s,q,r)\\text{–aloof}\\bigr\\}.\n\\]\n\n(a) Show that  \n\n\\[\nA_{n,k}^{(s,q,r)}\n=\\frac1q\\sum_{t=0}^{q-1}\\;\n\\omega^{-tr}\\,\n\\omega^{t\\,(s-1)k(k-1)/2}\\,\n\\omega^{t\\,k(k+1)/2}\\,\n\\bigl[{M\\choose k}\\bigr]_{\\omega^{t}},\\tag{$\\star$}\n\\]\n\nwhere  \n\n\\[\nM:=n-(s-1)(k-1),\\qquad \n\\omega:=e^{2\\pi i/q},\\qquad\n\\bigl[{M\\choose k}\\bigr]_{x}:=\n\\frac{(1-x^{M})(1-x^{M-1})\\cdots(1-x^{M-k+1})}\n     {(1-x)(1-x^{2})\\cdots(1-x^{k})}.\n\\]\n\n(Throughout we interpret the whole right–hand side as $0$ whenever $M<k$.)\n\n(b) Derive completely explicit expressions from $(\\star)$ in the two important special cases  \n\n(i) $q=2$ (parity restriction);  \n\n(ii) $s=2$ (classical “no two chosen positions are consecutive”).\n\n--------------------------------------------------------------------",
      "solution": "Throughout put  \n\n\\[\nM:=n-(s-1)(k-1).\\tag{1}\n\\]\n\nIf $M<k$ both sides of $(\\star)$ are $0$; hence we assume $M\\ge k$.\n\nStep 1.  Compress the spacing condition.  \nWrite $S=\\{x_{1}<\\dots <x_{k}\\}$ and set  \n\n\\[\ny_i:=x_i-(i-1)(s-1)\\qquad(1\\le i\\le k).\\tag{2}\n\\]\n\nBecause $x_{i+1}-x_i\\ge s$, the sequence $(y_i)$ is strictly increasing by at least $1$, so $\\{y_1,\\dots ,y_k\\}$ is an ordinary $k$–subset of $\\{1,\\dots ,M\\}$.  Conversely, every $k$–subset $\\{y_1<\\dots <y_k\\}\\subseteq\\{1,\\dots ,M\\}$ gives an $(s,q,r)$–aloof set via $x_i=y_i+(i-1)(s-1)$.  Thus (2) establishes a bijection.\n\nStep 2.  Translate the congruence.  \nSumming (2) yields  \n\n\\[\n\\sum_{i=1}^{k}x_i=\\sum_{i=1}^{k}y_i+(s-1)\\sum_{i=1}^{k}(i-1)\n                  =\\sum_{i=1}^{k}y_i+(s-1)\\,\\frac{k(k-1)}{2}.\\tag{3}\n\\]\n\nHence  \n\n\\[\n\\sum_{i=1}^{k}y_i\\equiv r-\\frac{(s-1)k(k-1)}{2}\\pmod q.\\tag{4}\n\\]\n\nPut  \n\n\\[\nc\\equiv r-\\frac{(s-1)k(k-1)}{2}\\pmod q.\\tag{5}\n\\]\n\nSo we must count $k$–subsets of $\\{1,\\dots ,M\\}$ whose element–sum lies in the residue class $c$ modulo $q$.\n\nStep 3.  A bivariate generating function and a root–of–unity filter.  \nIntroduce indeterminates $u$ (for the subset size) and $z$ (for the element–sum) and define  \n\n\\[\nF(u,z):=\\prod_{j=1}^{M}(1+u\\,z^{\\,j}).\\tag{6}\n\\]\n\nThe coefficient of $u^{k}z^{m}$ in $F$ counts $k$–subsets of $\\{1,\\dots ,M\\}$ with total sum $m$.  \nTo select the residue class $m\\equiv c\\pmod q$ we use the classical root–of–unity filter with $\\omega=e^{2\\pi i/q}$:\n\n\\[\nA_{n,k}^{(s,q,r)}\n=\\frac1q\\sum_{t=0}^{q-1}\\omega^{-tc}\\,\\bigl[u^{k}\\bigr]F\\!\\bigl(u,z=\\omega^{t}\\bigr).\\tag{7}\n\\]\n\nStep 4.  Extract $\\bigl[u^{k}\\bigr]F(u,\\omega^{t})$.  \nFix $t$ and abbreviate $\\xi:=\\omega^{t}$.  Then  \n\n\\[\nF(u,\\xi)=\\prod_{j=1}^{M}(1+u\\,\\xi^{\\,j})\n        =\\prod_{j=0}^{M-1}\\bigl(1+u\\,\\xi^{\\,j+1}\\bigr).\\tag{8}\n\\]\n\nPut $v:=u\\,\\xi$.  The $q$–binomial theorem (Gauss) reads\n\n\\[\n\\prod_{j=0}^{M-1}(1+y\\,x^{\\,j})\n     =\\sum_{\\ell=0}^{M}y^{\\ell}x^{\\,\\ell(\\ell-1)/2}\n       \\bigl[{M\\choose \\ell}\\bigr]_{x}.\\tag{9}\n\\]\n\nWith $y=v,\\;x=\\xi$ we get\n\n\\[\nF(u,\\xi)=\\sum_{\\ell=0}^{M}u^{\\ell}\\,\n          \\xi^{\\,\\ell(\\ell+1)/2}\\,\n          \\bigl[{M\\choose \\ell}\\bigr]_{\\xi}.\\tag{10}\n\\]\n\nHence  \n\n\\[\n\\bigl[u^{k}\\bigr]F(u,\\xi)=\\xi^{\\,k(k+1)/2}\\,\n                           \\bigl[{M\\choose k}\\bigr]_{\\xi}.\\tag{11}\n\\]\n\nStep 5.  Assembly.  \nInsert (11) into (7) and recall (5):\n\n\\[\n\\begin{aligned}\nA_{n,k}^{(s,q,r)}\n&=\\frac1q\\sum_{t=0}^{q-1}\n   \\omega^{-t\\,c}\\,\\omega^{\\,t\\,k(k+1)/2}\\,\n   \\bigl[{M\\choose k}\\bigr]_{\\omega^{t}} \\\\[2pt]\n&=\\frac1q\\sum_{t=0}^{q-1}\n   \\omega^{-tr}\\,\n   \\omega^{\\,t\\,(s-1)k(k-1)/2}\\,\n   \\omega^{\\,t\\,k(k+1)/2}\\,\n   \\bigl[{M\\choose k}\\bigr]_{\\omega^{t}},\n\\end{aligned}\n\\]\n\nwhich is exactly $(\\star)$.\n\n---------------------------------------------------------------  \nPart (b)  Specialisations.\n\n(i)  Modulus $q=2$.  \nHere $\\omega=-1$ and $t\\in\\{0,1\\}$.  Put  \n\n\\[\nE:=-r+\\frac{(s-1)k(k-1)}{2}+\\frac{k(k+1)}{2}.\\tag{12}\n\\]\n\nFrom $(\\star)$ we get  \n\n\\[\nA_{n,k}^{(s,2,r)}\n=\\frac12\\,\\binom{M}{k}\n+\\frac12\\,(-1)^{E}\\,\n          \\bigl[{M\\choose k}\\bigr]_{-1}.\\tag{13}\n\\]\n\nEvaluation of the Gaussian coefficient at $x=-1$.  \nWriting the $q$–binomial in product form and inspecting parities of the exponents gives the classical result\n\n\\[\n\\bigl[{M\\choose k}\\bigr]_{-1}\n=\\begin{cases}\n0, & k(M-k)\\text{ odd},\\\\[6pt]\n\\displaystyle\\binom{\\lfloor M/2\\rfloor}{\\lfloor k/2\\rfloor}, & k(M-k)\\text{ even}.\n\\end{cases}\\tag{14}\n\\]\n\nCombining (13) and (14) yields a completely explicit answer for all parity–restricted enumerations.\n\n(ii)  Separation $s=2$.  \nNow $M=n-k+1$, so $(\\star)$ becomes  \n\n\\[\nA_{n,k}^{(2,q,r)}\n=\\frac1q\\sum_{t=0}^{q-1}\n  \\omega^{\\,t\\,(k^{2}-r)}\\,\n  \\bigl[{\\,n-k+1\\choose k}\\bigr]_{\\omega^{t}}.\\tag{15}\n\\]\n\nTo make (15) fully explicit we expand the $q$–Gaussian into an ordinary **finite product** (no new notation involved):\n\n\\[\n\\bigl[{n-k+1\\choose k}\\bigr]_{\\omega^{t}}\n=\\prod_{j=1}^{k}\n  \\frac{1-\\omega^{t\\,(n-k+2-j)}}{1-\\omega^{t\\,j}}\\qquad(0\\le t\\le q-1).\\tag{16}\n\\]\n\nHence\n\n\\[\n\\boxed{\\;\nA_{n,k}^{(2,q,r)}\n=\\frac1q\\sum_{t=0}^{q-1}\\,\n  \\omega^{\\,t\\,(k^{2}-r)}\\;\n  \\prod_{j=1}^{k}\n     \\frac{1-\\omega^{t\\,(n-k+2-j)}}{1-\\omega^{t\\,j}}\n\\;}. \\tag{17}\n\\]\n\nEverything on the right–hand side of (17) is an *explicit* finite product of $2k$ factors, each factor a known $q$-th root of unity.  \nIn particular, no Gaussian symbol remains, and (17) is an entirely elementary closed form.\n\nRemark 1.  \nContrary to a claim in an earlier draft, the sum in (17) **never collapses to a single binomial coefficient** merely because $q\\mid k^{2}$; the root–of–unity filter\n$\\displaystyle \\sum_{t=0}^{q-1}\\omega^{t\\,(k^{2}-r)}(\\,\\cdots)$\ncontinues to interact with the product (16).  \nFor instance, with $(n,k,q,r)=(4,2,2,0)$ one finds\n$A_{4,2}^{(2,2,0)}=2\\neq\\binom{3}{2}$, in perfect agreement with (17).\n\nRemark 2 (optional refinement).  \nLet  \n\n\\[\nn-k+1=q\\,M_{1}+M_{0},\\qquad \nk      =q\\,k_{1}+k_{0}\\qquad(0\\le M_{0},k_{0}<q).\n\\]\n\nUsing the $q$–Lucas (or “\\(q\\)-adic”) factorisation of Gaussian coefficients at a primitive $q$-th root of unity one obtains\n\n\\[\n\\bigl[{n-k+1\\choose k}\\bigr]_{\\omega^{t}}=\n\\begin{cases}\n\\displaystyle(-1)^{t\\,\\Theta}\\,\n             \\binom{M_{1}}{k_{1}}\\,\n             \\binom{M_{0}}{k_{0}}, & k_{0}\\le M_{0},\\\\[8pt]\n0, & k_{0}>M_{0},\n\\end{cases}\\tag{18}\n\\]\n\nwhere $\\Theta$ is an explicit, easily written quadratic form in the digits $M_{0},k_{0}$.  \nInsertion of (18) into (15) yields a formula for $A_{n,k}^{(2,q,r)}$ that involves *only ordinary binomial coefficients* and elementary roots of unity; this is sometimes useful for concrete numerical work.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.402780",
        "was_fixed": false,
        "difficulty_analysis": "• Additional parameter s ≥ 2 forces a minimum spacing larger than 1, so the simple “delete the k−1 B’s” trick of the original solution no longer suffices; one must perform the more involved compression (1).  \n• An arithmetic congruence on the sum of the chosen indices introduces number-theoretic conditions absent from the original task.  \n• Solving the problem now needs three advanced tools:  \n 1. A stars-and-bars–type compression for s-spacing;  \n 2. q–analogue combinatorics (Gaussian binomials) to encode sums;  \n 3. A discrete Fourier / root–of–unity filter to isolate a specified residue class.  \n• The final closed form (★) intertwines graph-theoretic independence, additive number theory, and algebraic combinatorics; it cannot be obtained by elementary pattern matching or direct counting.  \n• Even the two “simpler” corollaries demand familiarity with parity arguments or with evaluating Gaussian coefficients at q = −1, again well beyond the original binomial-coefficient level.\n\nHence the enhanced kernel variant is substantially more technical and conceptually deeper than both the original problem and its current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}