{
  "index": "1987-B-2",
  "type": "COMB",
  "tag": [
    "COMB",
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "Let $r,s$ and $t$ be integers with $0 \\leq r$, $0 \\leq s$ and $r+s\n\\leq t$. Prove that\n\\[\n\\frac{\\binom s0}{\\binom tr}\n+ \\frac{\\binom s1}{\\binom{t}{r+1}} + \\cdots\n+ \\frac{\\binom ss}{\\binom{t}{r+s}}\n= \\frac{t+1}{(t+1-s)\\binom{t-s}{r}}.\n\\]",
  "solution": "Solution 1. We prove\n\\[\nF(r, s, t)=\\frac{t+1}{(t+1-s)\\binom{t-s}{r}}\n\\]\nby induction on \\( s \\). The base case \\( s=0 \\) is trivial since \\( \\binom{0}{0}=1 \\).\nFor \\( s \\geq 1, r \\geq 0 \\) and \\( r+s \\leq t \\),\n\\[\n\\begin{aligned}\nF(r, s, t) & =\\frac{\\binom{s-1}{0}}{\\binom{t}{r}}+\\frac{\\binom{s-1}{0}+\\binom{s-1}{1}}{\\binom{t}{r+1}}+\\cdots+\\frac{\\binom{s-1}{s-2}+\\binom{s-1}{s-1}}{\\binom{t}{r+s-1}}+\\frac{\\binom{s-1}{s-1}}{\\binom{t}{r+s}} \\\\\n& =F(r, s-1, t)+F(r+1, s-1, t) .\n\\end{aligned}\n\\]\n\nApplying the inductive hypothesis to the two terms on the right gives\n\\[\nF(r, s, t)=\\frac{t+1}{(t+2-s)\\binom{t+1-s}{r}}+\\frac{t+1}{(t+2-s)\\binom{t+1-s}{r+1}} .\n\\]\n\nThe definition of binomial coefficients in terms of factorials lets us express \\( \\binom{t+1-s}{r} \\) and \\( \\binom{t+1-s}{r+1} \\) in terms of \\( \\binom{t-s}{r} \\); this leads to\n\\[\n\\begin{aligned}\nF(r, s, t) & =\\frac{t+1}{(t+2-s)\\left(\\frac{t+1-s}{t+1-s-r}\\right)\\binom{t-s}{r}}+\\frac{t+1}{(t+2-s)\\left(\\frac{t+1-s}{r+1}\\right)\\binom{t-s}{r}} \\\\\n& =\\frac{t+1}{(t+2-s)\\binom{t-s}{r}}\\left(\\frac{t+1-s-r}{t+1-s}+\\frac{r+1}{t+1-s}\\right) \\\\\n& =\\frac{t+1\\binom{t-s}{r}}{(t+1-s)}\n\\end{aligned}\n\\]\ncompleting the inductive step.\nSolution 2. Writing the binomial coefficients in terms of factorials and regrouping, we find\n\\[\nF(r, s, t)=\\frac{s!r!(t-r-s)!}{t!} \\sum_{i=0}^{s}\\binom{r+i}{r}\\binom{t-r-i}{t-r-s} .\n\\]\n\nIf we could prove\n\\[\n\\sum_{i=0}^{s}\\binom{r+i}{r}\\binom{t-r-i}{t-r-s}=\\binom{t+1}{t-s+1}\n\\]\nthen substituting into (1) would yield\n\\[\nF(r, s, t)=\\frac{s!r!(t-r-s)!}{t!} \\cdot \\frac{(t+1)!}{(t-s+1)!s!}=\\frac{t+1}{(t+1-s)\\binom{t-s}{r}} .\n\\]\n\nWe now provide three proofs of (2):\nProof 1 (Vandermonde's identity). We have\n\\[\n\\begin{aligned}\n\\binom{r+i}{r} & =\\binom{r+i}{i} \\\\\n& =\\frac{(r+i)(r+i-1) \\cdots(r+1)}{i!} \\\\\n& =(-1)^{i} \\frac{(-r-1)(-r-2) \\cdots(-r-i)}{i!} \\\\\n& =(-1)^{i}\\binom{-r-1}{i}\n\\end{aligned}\n\\]\nand similarly\n\\[\n\\binom{t-r-i}{t-r-s}=\\binom{(t-r-s)+(s-i)}{s-i}=\\cdots=(-1)^{s-i}\\binom{-t+r+s-1}{s-i}\n\\]\nand\n\\[\n\\binom{t+1}{t-s+1}=\\binom{(t-s+1)+s}{s}=\\cdots=(-1)^{s}\\binom{-t+s-2}{s} .\n\\]\n\nThus (2) can be rewritten as\n\\[\n\\sum_{i=0}^{s}\\binom{-r-1}{i}\\binom{-t+r+s-1}{s-i}=\\binom{-t+s-2}{s}\n\\]\nby multiplying both sides by \\( (-1)^{s} \\). This is a special case of Vandermonde's identity, which in general states that for integers \\( m, n, s \\) with \\( s \\geq 0 \\),\n\\[\n\\sum_{i=0}^{s}\\binom{m}{i}\\binom{n}{s-i}=\\binom{m+n}{s}\n\\]\n\nProof 2 (generating functions). The binomial expansion gives\n\\[\n(1-x)^{-(n+1)}=\\sum_{i=0}^{\\infty}\\binom{-n-1}{i}(-x)^{i}=\\sum_{i=0}^{\\infty}\\binom{n+i}{n} x^{i}\n\\]\nso taking coefficients of \\( x^{s} \\) in the identity \\( (1-x)^{-(r+1)}(1-x)^{-(t-r-s+1)}= \\) \\( (1-x)^{-(t-s+2)} \\) yields (2).\n\nProof 3 (bijective proof). We will show that the two sides of (2) count something in two different ways. First set \\( j=r+i \\) to rewrite (2) as\n\\[\n\\sum_{j=r}^{r+s}\\binom{j}{r}\\binom{t-j}{t-r-s}=\\binom{t+1}{s}\n\\]\n\nWe will show that both sides of (3) count the number of sequences of \\( s \\) zeros and \\( t-s \\) ones, punctuated by a comma such that the number of ones occurring before the comma is \\( r \\). On one hand, the number of such sequences of \\( t+1 \\) symbols (including the comma) equals the right-hand side \\( \\binom{t+1}{s} \\) of (3), because they can be constructed by choosing the \\( s \\) positions for the zeros: of the remaining positions, the \\( (r+1)^{\\text {st }} \\) must contain the comma and the others must contain ones. On the other hand, we can count the sequences according to the position of the comma: given that there are exactly \\( j \\) digits before the comma, there are \\( \\binom{j}{r} \\) possibilities for the digits before the comma (since \\( r \\) of them are to be ones and the rest are to be zeros), and \\( \\binom{t-j}{t-r-s} \\) possibilities for the digits after the comma (since one needs \\( t-r-s \\) more ones to bring the total number of ones to \\( t-s \\) ). Summing over \\( j \\) shows that the total number of sequences is the left-hand side of (3).\n\nMotivation. To find the generating function solution, first observe that the sum in (1) looks like the coefficient of \\( x^{s} \\) in a product of two series, and then figure out what the two series must be.\n\nRemark. Vandermonde's identity can also be used in 1991B4.\nLiterature note. Generating functions are a powerful method for proving combinatorial identities. A comprehensive introduction to this method is [Wi].\n\nRelated question. Problem 20 of \\( [\\mathrm{WH}] \\) is similar:\nEvaluate the sum\n\\[\nS=\\sum_{k=0}^{n} \\frac{\\binom{n}{k}}{\\binom{2 n-1}{k}}\n\\]\nfor all positive integers \\( n \\).\n(Hint: the answer does not depend on \\( n \\).)",
  "vars": [
    "i",
    "j",
    "k",
    "x",
    "F",
    "S"
  ],
  "params": [
    "r",
    "s",
    "t",
    "n",
    "m"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "i": "indexvar",
        "j": "positionvar",
        "k": "iteratorvar",
        "x": "placeholder",
        "F": "functionf",
        "S": "summationvar",
        "r": "countones",
        "s": "countzeros",
        "t": "totalsize",
        "n": "integern",
        "m": "integerm"
      },
      "question": "Let $countones,countzeros$ and $totalsize$ be integers with $0 \\leq countones$, $0 \\leq countzeros$ and $countones+countzeros \\leq totalsize$. Prove that\n\\[\n\\frac{\\binom{countzeros}{0}}{\\binom{totalsize}{countones}}\n+ \\frac{\\binom{countzeros}{1}}{\\binom{totalsize}{countones+1}} + \\cdots\n+ \\frac{\\binom{countzeros}{countzeros}}{\\binom{totalsize}{countones+countzeros}}\n= \\frac{totalsize+1}{(totalsize+1-countzeros)\\binom{totalsize-countzeros}{countones}}.\n\\]",
      "solution": "Solution 1. We prove\n\\[\nfunctionf(countones, countzeros, totalsize)=\\frac{totalsize+1}{(totalsize+1-countzeros)\\binom{totalsize-countzeros}{countones}}\n\\]\nby induction on \\( countzeros \\). The base case \\( countzeros=0 \\) is trivial since \\( \\binom{0}{0}=1 \\).\nFor \\( countzeros \\geq 1, countones \\geq 0 \\) and \\( countones+countzeros \\leq totalsize \\),\n\\[\n\\begin{aligned}\nfunctionf(countones, countzeros, totalsize) & =\\frac{\\binom{countzeros-1}{0}}{\\binom{totalsize}{countones}}+\\frac{\\binom{countzeros-1}{0}+\\binom{countzeros-1}{1}}{\\binom{totalsize}{countones+1}}+\\cdots+\\frac{\\binom{countzeros-1}{countzeros-2}+\\binom{countzeros-1}{countzeros-1}}{\\binom{totalsize}{countones+countzeros-1}}+\\frac{\\binom{countzeros-1}{countzeros-1}}{\\binom{totalsize}{countones+countzeros}} \\\\\n& =functionf(countones, countzeros-1, totalsize)+functionf(countones+1, countzeros-1, totalsize) .\n\\end{aligned}\n\\]\n\nApplying the inductive hypothesis to the two terms on the right gives\n\\[\nfunctionf(countones, countzeros, totalsize)=\\frac{totalsize+1}{(totalsize+2-countzeros)\\binom{totalsize+1-countzeros}{countones}}+\\frac{totalsize+1}{(totalsize+2-countzeros)\\binom{totalsize+1-countzeros}{countones+1}} .\n\\]\n\nThe definition of binomial coefficients in terms of factorials lets us express \\( \\binom{totalsize+1-countzeros}{countones} \\) and \\( \\binom{totalsize+1-countzeros}{countones+1} \\) in terms of \\( \\binom{totalsize-countzeros}{countones} \\); this leads to\n\\[\n\\begin{aligned}\nfunctionf(countones, countzeros, totalsize) & =\\frac{totalsize+1}{(totalsize+2-countzeros)\\left(\\frac{totalsize+1-countzeros}{totalsize+1-countzeros-countones}\\right)\\binom{totalsize-countzeros}{countones}}+\\frac{totalsize+1}{(totalsize+2-countzeros)\\left(\\frac{totalsize+1-countzeros}{countones+1}\\right)\\binom{totalsize-countzeros}{countones}} \\\\\n& =\\frac{totalsize+1}{(totalsize+2-countzeros)\\binom{totalsize-countzeros}{countones}}\\left(\\frac{totalsize+1-countzeros-countones}{totalsize+1-countzeros}+\\frac{countones+1}{totalsize+1-countzeros}\\right) \\\\\n& =\\frac{totalsize+1}{(totalsize+1-countzeros)\\binom{totalsize-countzeros}{countones}},\n\\end{aligned}\n\\]\ncompleting the inductive step.\n\nSolution 2. Writing the binomial coefficients in terms of factorials and regrouping, we find\n\\[\nfunctionf(countones, countzeros, totalsize)=\\frac{countzeros!\\,countones!\\,(totalsize-countones-countzeros)!}{totalsize!} \\sum_{indexvar=0}^{countzeros}\\binom{countones+indexvar}{countones}\\binom{totalsize-countones-indexvar}{totalsize-countones-countzeros} .\n\\tag{1}\n\\]\n\nIf we could prove\n\\[\n\\sum_{indexvar=0}^{countzeros}\\binom{countones+indexvar}{countones}\\binom{totalsize-countones-indexvar}{totalsize-countones-countzeros}=\\binom{totalsize+1}{totalsize-countzeros+1}\n\\tag{2}\n\\]\nthen substituting into (1) would yield\n\\[\nfunctionf(countones, countzeros, totalsize)=\\frac{countzeros!\\,countones!\\,(totalsize-countones-countzeros)!}{totalsize!} \\cdot \\frac{(totalsize+1)!}{(totalsize-countzeros+1)!countzeros!}=\\frac{totalsize+1}{(totalsize+1-countzeros)\\binom{totalsize-countzeros}{countones}} .\n\\]\n\nWe now provide three proofs of (2):\n\nProof 1 (Vandermonde's identity). We have\n\\[\n\\begin{aligned}\n\\binom{countones+indexvar}{countones} & =\\binom{countones+indexvar}{indexvar} \\\\\n& =\\frac{(countones+indexvar)(countones+indexvar-1) \\cdots(countones+1)}{indexvar!} \\\\\n& =(-1)^{indexvar} \\frac{(-countones-1)(-countones-2) \\cdots(-countones-indexvar)}{indexvar!} \\\\\n& =(-1)^{indexvar}\\binom{-countones-1}{indexvar}\n\\end{aligned}\n\\]\nand similarly\n\\[\n\\binom{totalsize-countones-indexvar}{totalsize-countones-countzeros}=\\binom{(totalsize-countones-countzeros)+(countzeros-indexvar)}{countzeros-indexvar}=\\cdots=(-1)^{countzeros-indexvar}\\binom{-totalsize+countones+countzeros-1}{countzeros-indexvar}\n\\]\nand\n\\[\n\\binom{totalsize+1}{totalsize-countzeros+1}=\\binom{(totalsize-countzeros+1)+countzeros}{countzeros}=\\cdots=(-1)^{countzeros}\\binom{-totalsize+countzeros-2}{countzeros} .\n\\]\n\nThus (2) can be rewritten as\n\\[\n\\sum_{indexvar=0}^{countzeros}\\binom{-countones-1}{indexvar}\\binom{-totalsize+countones+countzeros-1}{countzeros-indexvar}=\\binom{-totalsize+countzeros-2}{countzeros}\n\\]\nby multiplying both sides by \\( (-1)^{countzeros} \\). This is a special case of Vandermonde's identity, which in general states that for integers \\( integerm, integern, countzeros \\) with \\( countzeros \\geq 0 \\),\n\\[\n\\sum_{indexvar=0}^{countzeros}\\binom{integerm}{indexvar}\\binom{integern}{countzeros-indexvar}=\\binom{integerm+integern}{countzeros}.\n\\]\n\nProof 2 (generating functions). The binomial expansion gives\n\\[\n(1-placeholder)^{-(integern+1)}=\\sum_{indexvar=0}^{\\infty}\\binom{-integern-1}{indexvar}(-placeholder)^{indexvar}=\\sum_{indexvar=0}^{\\infty}\\binom{integern+indexvar}{integern} placeholder^{indexvar},\n\\]\nso taking coefficients of \\( placeholder^{countzeros} \\) in the identity \\( (1-placeholder)^{-(countones+1)}(1-placeholder)^{-(totalsize-countones-countzeros+1)}=(1-placeholder)^{-(totalsize-countzeros+2)} \\) yields (2).\n\nProof 3 (bijective proof). We will show that the two sides of (2) count something in two different ways. First set \\( positionvar=countones+indexvar \\) to rewrite (2) as\n\\[\n\\sum_{positionvar=countones}^{countones+countzeros}\\binom{positionvar}{countones}\\binom{totalsize-positionvar}{totalsize-countones-countzeros}=\\binom{totalsize+1}{countzeros}.\n\\]\n\nWe will show that both sides of (3) count the number of sequences of \\( countzeros \\) zeros and \\( totalsize-countzeros \\) ones, punctuated by a comma such that the number of ones occurring before the comma is \\( countones \\). On one hand, the number of such sequences of \\( totalsize+1 \\) symbols (including the comma) equals the right-hand side \\( \\binom{totalsize+1}{countzeros} \\) of (3), because they can be constructed by choosing the \\( countzeros \\) positions for the zeros: of the remaining positions, the \\( (countones+1)^{\\text {st}} \\) must contain the comma and the others must contain ones. On the other hand, we can count the sequences according to the position of the comma: given that there are exactly \\( positionvar \\) digits before the comma, there are \\( \\binom{positionvar}{countones} \\) possibilities for the digits before the comma (since \\( countones \\) of them are to be ones and the rest are to be zeros), and \\( \\binom{totalsize-positionvar}{totalsize-countones-countzeros} \\) possibilities for the digits after the comma (since one needs \\( totalsize-countones-countzeros \\) more ones to bring the total number of ones to \\( totalsize-countzeros \\) ). Summing over \\( positionvar \\) shows that the total number of sequences is the left-hand side of (3).\n\nMotivation. To find the generating function solution, first observe that the sum in (1) looks like the coefficient of \\( placeholder^{countzeros} \\) in a product of two series, and then figure out what the two series must be.\n\nRemark. Vandermonde's identity can also be used in 1991B4.\n\nLiterature note. Generating functions are a powerful method for proving combinatorial identities. A comprehensive introduction to this method is [Wi].\n\nRelated question. Problem 20 of \\( [\\mathrm{WH}] \\) is similar:\nEvaluate the sum\n\\[\nsummationvar=\\sum_{iteratorvar=0}^{integern} \\frac{\\binom{integern}{iteratorvar}}{\\binom{2\\,integern-1}{iteratorvar}}\n\\]\nfor all positive integers \\( integern \\).\n(Hint: the answer does not depend on \\( integern \\).)"
    },
    "descriptive_long_confusing": {
      "map": {
        "i": "firestone",
        "j": "goldcrest",
        "k": "hearthrug",
        "x": "inkblotter",
        "F": "journeyman",
        "S": "kingfisher",
        "r": "almondtree",
        "s": "blueprint",
        "t": "candelabra",
        "n": "driftwood",
        "m": "eggbeater"
      },
      "question": "Let $almondtree, blueprint$ and $candelabra$ be integers with $0 \\leq almondtree$, $0 \\leq blueprint$ and $almondtree+blueprint\\leq candelabra$. Prove that\n\\[\n\\frac{\\binom{blueprint}{0}}{\\binom{candelabra}{almondtree}}+ \\frac{\\binom{blueprint}{1}}{\\binom{candelabra}{almondtree+1}} + \\cdots+ \\frac{\\binom{blueprint}{blueprint}}{\\binom{candelabra}{almondtree+blueprint}}= \\frac{candelabra+1}{(candelabra+1-blueprint)\\binom{candelabra-blueprint}{almondtree}}.\n\\]",
      "solution": "Solution 1. We prove\n\\[\njourneyman(almondtree, blueprint, candelabra)=\\frac{candelabra+1}{(candelabra+1-blueprint)\\binom{candelabra-blueprint}{almondtree}}\n\\]\nby induction on $ blueprint $. The base case $ blueprint=0 $ is trivial since $ \\binom{0}{0}=1 $. For $ blueprint \\geq 1, almondtree \\geq 0 $ and $ almondtree+blueprint \\leq candelabra $,\n\\[\n\\begin{aligned}\njourneyman(almondtree, blueprint, candelabra) &=\\frac{\\binom{blueprint-1}{0}}{\\binom{candelabra}{almondtree}}+\\frac{\\binom{blueprint-1}{0}+\\binom{blueprint-1}{1}}{\\binom{candelabra}{almondtree+1}}+\\cdots+\\frac{\\binom{blueprint-1}{blueprint-2}+\\binom{blueprint-1}{blueprint-1}}{\\binom{candelabra}{almondtree+blueprint-1}}+\\frac{\\binom{blueprint-1}{blueprint-1}}{\\binom{candelabra}{almondtree+blueprint}}\\\\\n&=journeyman(almondtree, blueprint-1, candelabra)+journeyman(almondtree+1, blueprint-1, candelabra).\n\\end{aligned}\n\\]\nApplying the inductive hypothesis to the two terms on the right gives\n\\[\njourneyman(almondtree, blueprint, candelabra)=\\frac{candelabra+1}{(candelabra+2-blueprint)\\binom{candelabra+1-blueprint}{almondtree}}+\\frac{candelabra+1}{(candelabra+2-blueprint)\\binom{candelabra+1-blueprint}{almondtree+1}}.\n\\]\nThe definition of binomial coefficients in terms of factorials lets us express $ \\binom{candelabra+1-blueprint}{almondtree} $ and $ \\binom{candelabra+1-blueprint}{almondtree+1} $ in terms of $ \\binom{candelabra-blueprint}{almondtree} $; this leads to\n\\[\n\\begin{aligned}\njourneyman(almondtree, blueprint, candelabra) &=\\frac{candelabra+1}{(candelabra+2-blueprint)\\left(\\frac{candelabra+1-blueprint}{candelabra+1-blueprint-almondtree}\\right)\\binom{candelabra-blueprint}{almondtree}}+\\frac{candelabra+1}{(candelabra+2-blueprint)\\left(\\frac{candelabra+1-blueprint}{almondtree+1}\\right)\\binom{candelabra-blueprint}{almondtree}}\\\\\n&=\\frac{candelabra+1}{(candelabra+2-blueprint)\\binom{candelabra-blueprint}{almondtree}}\\left(\\frac{candelabra+1-blueprint-almondtree}{candelabra+1-blueprint}+\\frac{almondtree+1}{candelabra+1-blueprint}\\right)\\\\\n&=\\frac{candelabra+1}{(candelabra+1-blueprint)\\binom{candelabra-blueprint}{almondtree}}\n\\end{aligned}\n\\]\ncompleting the inductive step.\n\nSolution 2. Writing the binomial coefficients in terms of factorials and regrouping, we find\n\\[\njourneyman(almondtree, blueprint, candelabra)=\\frac{blueprint!\\,almondtree!\\,(candelabra-almondtree-blueprint)!}{candelabra!}\\sum_{firestone=0}^{blueprint}\\binom{almondtree+firestone}{almondtree}\\binom{candelabra-almondtree-firestone}{candelabra-almondtree-blueprint}.\n\\]\nIf we could prove\n\\[\n\\sum_{firestone=0}^{blueprint}\\binom{almondtree+firestone}{almondtree}\\binom{candelabra-almondtree-firestone}{candelabra-almondtree-blueprint}=\\binom{candelabra+1}{candelabra-blueprint+1}\n\\]\nthen substituting into (1) would yield\n\\[\njourneyman(almondtree, blueprint, candelabra)=\\frac{blueprint!\\,almondtree!\\,(candelabra-almondtree-blueprint)!}{candelabra!}\\cdot\\frac{(candelabra+1)!}{(candelabra-blueprint+1)!\\,blueprint!}=\\frac{candelabra+1}{(candelabra+1-blueprint)\\binom{candelabra-blueprint}{almondtree}}.\n\\]\n\nWe now provide three proofs of (2):\n\nProof 1 (Vandermonde's identity). We have\n\\[\n\\begin{aligned}\n\\binom{almondtree+firestone}{almondtree}&=\\binom{almondtree+firestone}{firestone}=\\frac{(almondtree+firestone)(almondtree+firestone-1)\\cdots(almondtree+1)}{firestone!}=(-1)^{firestone}\\frac{(-almondtree-1)(-almondtree-2)\\cdots(-almondtree-firestone)}{firestone!}=(-1)^{firestone}\\binom{-almondtree-1}{firestone}\n\\end{aligned}\n\\]\nand similarly\n\\[\n\\binom{candelabra-almondtree-firestone}{candelabra-almondtree-blueprint}=\\binom{(candelabra-almondtree-blueprint)+(blueprint-firestone)}{blueprint-firestone}=\\cdots=(-1)^{blueprint-firestone}\\binom{-candelabra+almondtree+blueprint-1}{blueprint-firestone}\n\\]\nand\n\\[\n\\binom{candelabra+1}{candelabra-blueprint+1}=\\binom{(candelabra-blueprint+1)+blueprint}{blueprint}=\\cdots=(-1)^{blueprint}\\binom{-candelabra+blueprint-2}{blueprint}.\n\\]\nThus (2) can be rewritten as\n\\[\n\\sum_{firestone=0}^{blueprint}\\binom{-almondtree-1}{firestone}\\binom{-candelabra+almondtree+blueprint-1}{blueprint-firestone}=\\binom{-candelabra+blueprint-2}{blueprint}\n\\]\nby multiplying both sides by $ (-1)^{blueprint} $. This is a special case of Vandermonde's identity, which in general states that for integers $ eggbeater, driftwood, blueprint $ with $ blueprint \\geq 0 $,\n\\[\n\\sum_{firestone=0}^{blueprint}\\binom{eggbeater}{firestone}\\binom{driftwood}{blueprint-firestone}=\\binom{eggbeater+driftwood}{blueprint}.\n\\]\n\nProof 2 (generating functions). The binomial expansion gives\n\\[\n(1-inkblotter)^{-(driftwood+1)}=\\sum_{firestone=0}^{\\infty}\\binom{-driftwood-1}{firestone}(-inkblotter)^{firestone}=\\sum_{firestone=0}^{\\infty}\\binom{driftwood+firestone}{driftwood}inkblotter^{firestone},\n\\]\nso taking coefficients of $ inkblotter^{blueprint} $ in the identity $ (1-inkblotter)^{-(almondtree+1)}(1-inkblotter)^{-(candelabra-almondtree-blueprint+1)}=(1-inkblotter)^{-(candelabra-blueprint+2)} $ yields (2).\n\nProof 3 (bijective proof). We will show that the two sides of (2) count something in two different ways. First set $ goldcrest=almondtree+firestone $ to rewrite (2) as\n\\[\n\\sum_{goldcrest=almondtree}^{almondtree+blueprint}\\binom{goldcrest}{almondtree}\\binom{candelabra-goldcrest}{candelabra-almondtree-blueprint}=\\binom{candelabra+1}{blueprint}.\n\\]\nWe will show that both sides of (3) count the number of sequences of $ blueprint $ zeros and $ candelabra-blueprint $ ones, punctuated by a comma such that the number of ones occurring before the comma is $ almondtree $. On one hand, the number of such sequences of $ candelabra+1 $ symbols (including the comma) equals the right-hand side $ \\binom{candelabra+1}{blueprint} $ of (3), because they can be constructed by choosing the $ blueprint $ positions for the zeros: of the remaining positions, the $ (almondtree+1)^{\\text {st }} $ must contain the comma and the others must contain ones. On the other hand, we can count the sequences according to the position of the comma: given that there are exactly $ goldcrest $ digits before the comma, there are $ \\binom{goldcrest}{almondtree} $ possibilities for the digits before the comma (since $ almondtree $ of them are to be ones and the rest are to be zeros), and $ \\binom{candelabra-goldcrest}{candelabra-almondtree-blueprint} $ possibilities for the digits after the comma (since one needs $ candelabra-almondtree-blueprint $ more ones to bring the total number of ones to $ candelabra-blueprint $). Summing over $ goldcrest $ shows that the total number of sequences is the left-hand side of (3).\n\nMotivation. To find the generating function solution, first observe that the sum in (1) looks like the coefficient of $ inkblotter^{blueprint} $ in a product of two series, and then figure out what the two series must be.\n\nRemark. Vandermonde's identity can also be used in 1991B4.\n\nLiterature note. Generating functions are a powerful method for proving combinatorial identities. A comprehensive introduction to this method is [Wi].\n\nRelated question. Problem 20 of $ [\\mathrm{WH}] $ is similar: Evaluate the sum\n\\[\nkingfisher=\\sum_{hearthrug=0}^{driftwood} \\frac{\\binom{driftwood}{hearthrug}}{\\binom{2 driftwood-1}{hearthrug}}\n\\]\nfor all positive integers $ driftwood $. (Hint: the answer does not depend on $ driftwood $.)"
    },
    "descriptive_long_misleading": {
      "map": {
        "i": "realvalue",
        "j": "constant",
        "k": "stationary",
        "x": "knownvalue",
        "F": "argument",
        "S": "contrast",
        "r": "perimeter",
        "s": "difference",
        "t": "distance",
        "n": "positive",
        "m": "maximums"
      },
      "question": "Let $perimeter,difference$ and $distance$ be integers with $0 \\leq perimeter$, $0 \\leq difference$ and $perimeter+difference \\leq distance$. Prove that\n\\[\n\\frac{\\binom difference0}{\\binom distanceperimeter}\n+ \\frac{\\binom difference1}{\\binom{distance}{perimeter+1}} + \\cdots\n+ \\frac{\\binom difference difference}{\\binom{distance}{perimeter+difference}}\n= \\frac{distance+1}{(distance+1-difference)\\binom{distance-difference}{perimeter}}.\n\\]",
      "solution": "Solution 1. We prove\n\\[\nargument(perimeter, difference, distance)=\\frac{distance+1}{(distance+1-difference)\\binom{distance-difference}{perimeter}}\n\\]\nby induction on \\( difference \\). The base case \\( difference=0 \\) is trivial since \\( \\binom{0}{0}=1 \\).\nFor \\( difference \\geq 1, perimeter \\geq 0 \\) and \\( perimeter+difference \\leq distance \\),\n\\[\n\\begin{aligned}\nargument(perimeter, difference, distance) & =\\frac{\\binom{difference-1}{0}}{\\binom{distance}{perimeter}}+\\frac{\\binom{difference-1}{0}+\\binom{difference-1}{1}}{\\binom{distance}{perimeter+1}}+\\cdots+\\frac{\\binom{difference-1}{difference-2}+\\binom{difference-1}{difference-1}}{\\binom{distance}{perimeter+difference-1}}+\\frac{\\binom{difference-1}{difference-1}}{\\binom{distance}{perimeter+difference}} \\\\\n& =argument(perimeter, difference-1, distance)+argument(perimeter+1, difference-1, distance) .\n\\end{aligned}\n\\]\n\nApplying the inductive hypothesis to the two terms on the right gives\n\\[\nargument(perimeter, difference, distance)=\\frac{distance+1}{(distance+2-difference)\\binom{distance+1-difference}{perimeter}}+\\frac{distance+1}{(distance+2-difference)\\binom{distance+1-difference}{perimeter+1}} .\n\\]\n\nThe definition of binomial coefficients in terms of factorials lets us express \\( \\binom{distance+1-difference}{perimeter} \\) and \\( \\binom{distance+1-difference}{perimeter+1} \\) in terms of \\( \\binom{distance-difference}{perimeter} \\); this leads to\n\\[\n\\begin{aligned}\nargument(perimeter, difference, distance) & =\\frac{distance+1}{(distance+2-difference)\\left(\\frac{distance+1-difference}{distance+1-difference-perimeter}\\right)\\binom{distance-difference}{perimeter}}+\\frac{distance+1}{(distance+2-difference)\\left(\\frac{distance+1-difference}{perimeter+1}\\right)\\binom{distance-difference}{perimeter}} \\\\\n& =\\frac{distance+1}{(distance+2-difference)\\binom{distance-difference}{perimeter}}\\left(\\frac{distance+1-difference-perimeter}{distance+1-difference}+\\frac{perimeter+1}{distance+1-difference}\\right) \\\\\n& =\\frac{distance+1\\binom{distance-difference}{perimeter}}{(distance+1-difference)}\n\\end{aligned}\n\\]\ncompleting the inductive step.\n\nSolution 2. Writing the binomial coefficients in terms of factorials and regrouping, we find\n\\[\nargument(perimeter, difference, distance)=\\frac{difference!perimeter!(distance-perimeter-difference)!}{distance!} \\sum_{realvalue=0}^{difference}\\binom{perimeter+realvalue}{perimeter}\\binom{distance-perimeter-realvalue}{distance-perimeter-difference} .\n\\]\n\nIf we could prove\n\\[\n\\sum_{realvalue=0}^{difference}\\binom{perimeter+realvalue}{perimeter}\\binom{distance-perimeter-realvalue}{distance-perimeter-difference}=\\binom{distance+1}{distance-difference+1}\n\\]\nthen substituting into (1) would yield\n\\[\nargument(perimeter, difference, distance)=\\frac{difference!perimeter!(distance-perimeter-difference)!}{distance!} \\cdot \\frac{(distance+1)!}{(distance-difference+1)!difference!}=\\frac{distance+1}{(distance+1-difference)\\binom{distance-difference}{perimeter}} .\n\\]\n\nWe now provide three proofs of (2):\nProof 1 (Vandermonde's identity). We have\n\\[\n\\begin{aligned}\n\\binom{perimeter+realvalue}{perimeter} & =\\binom{perimeter+realvalue}{realvalue} \\\\\n& =\\frac{(perimeter+realvalue)(perimeter+realvalue-1) \\cdots(perimeter+1)}{realvalue!} \\\\\n& =(-1)^{realvalue} \\frac{(-perimeter-1)(-perimeter-2) \\cdots(-perimeter-realvalue)}{realvalue!} \\\\\n& =(-1)^{realvalue}\\binom{-perimeter-1}{realvalue}\n\\end{aligned}\n\\]\nand similarly\n\\[\n\\binom{distance-perimeter-realvalue}{distance-perimeter-difference}=\\binom{(distance-perimeter-difference)+(difference-realvalue)}{difference-realvalue}=\\cdots=(-1)^{difference-realvalue}\\binom{-distance+perimeter+difference-1}{difference-realvalue}\n\\]\nand\n\\[\n\\binom{distance+1}{distance-difference+1}=\\binom{(distance-difference+1)+difference}{difference}=\\cdots=(-1)^{difference}\\binom{-distance+difference-2}{difference} .\n\\]\n\nThus (2) can be rewritten as\n\\[\n\\sum_{realvalue=0}^{difference}\\binom{-perimeter-1}{realvalue}\\binom{-distance+perimeter+difference-1}{difference-realvalue}=\\binom{-distance+difference-2}{difference}\n\\]\nby multiplying both sides by \\( (-1)^{difference} \\). This is a special case of Vandermonde's identity, which in general states that for integers \\( maximums, positive, difference \\) with \\( difference \\geq 0 \\),\n\\[\n\\sum_{realvalue=0}^{difference}\\binom{maximums}{realvalue}\\binom{positive}{difference-realvalue}=\\binom{maximums+positive}{difference}\n\\]\n\nProof 2 (generating functions). The binomial expansion gives\n\\[\n(1-knownvalue)^{-(positive+1)}=\\sum_{realvalue=0}^{\\infty}\\binom{-positive-1}{realvalue}(-knownvalue)^{realvalue}=\\sum_{realvalue=0}^{\\infty}\\binom{positive+realvalue}{positive} knownvalue^{realvalue}\n\\]\nso taking coefficients of \\( knownvalue^{difference} \\) in the identity \\( (1-knownvalue)^{-(perimeter+1)}(1-knownvalue)^{-(distance-perimeter-difference+1)}= (1-knownvalue)^{-(distance-difference+2)} \\) yields (2).\n\nProof 3 (bijective proof). We will show that the two sides of (2) count something in two different ways. First set \\( constant=perimeter+realvalue \\) to rewrite (2) as\n\\[\n\\sum_{constant=perimeter}^{perimeter+difference}\\binom{constant}{perimeter}\\binom{distance-constant}{distance-perimeter-difference}=\\binom{distance+1}{difference}\n\\]\n\nWe will show that both sides of (3) count the number of sequences of \\( difference \\) zeros and \\( distance-difference \\) ones, punctuated by a comma such that the number of ones occurring before the comma is \\( perimeter \\). On one hand, the number of such sequences of \\( distance+1 \\) symbols (including the comma) equals the right-hand side \\( \\binom{distance+1}{difference} \\) of (3), because they can be constructed by choosing the \\( difference \\) positions for the zeros: of the remaining positions, the \\( (perimeter+1)^{\\text {st }} \\) must contain the comma and the others must contain ones. On the other hand, we can count the sequences according to the position of the comma: given that there are exactly \\( constant \\) digits before the comma, there are \\( \\binom{constant}{perimeter} \\) possibilities for the digits before the comma (since \\( perimeter \\) of them are to be ones and the rest are to be zeros), and \\( \\binom{distance-constant}{distance-perimeter-difference} \\) possibilities for the digits after the comma (since one needs \\( distance-perimeter-difference \\) more ones to bring the total number of ones to \\( distance-difference \\) ). Summing over \\( constant \\) shows that the total number of sequences is the left-hand side of (3).\n\nMotivation. To find the generating function solution, first observe that the sum in (1) looks like the coefficient of \\( knownvalue^{difference} \\) in a product of two series, and then figure out what the two series must be.\n\nRemark. Vandermonde's identity can also be used in 1991B4.\n\nLiterature note. Generating functions are a powerful method for proving combinatorial identities. A comprehensive introduction to this method is [Wi].\n\nRelated question. Problem 20 of \\( [\\mathrm{WH}] \\) is similar:\nEvaluate the sum\n\\[\ncontrast=\\sum_{stationary=0}^{positive} \\frac{\\binom{positive}{stationary}}{\\binom{2 positive-1}{stationary}}\n\\]\nfor all positive integers \\( positive \\).\n(Hint: the answer does not depend on \\( positive \\).)"
    },
    "garbled_string": {
      "map": {
        "i": "qbvrjkdu",
        "j": "zdplmyna",
        "k": "rscwfeoh",
        "x": "ulkpgzsa",
        "F": "hsqtnmve",
        "S": "jdrwqkoi",
        "r": "pczmyeor",
        "s": "gvnhudab",
        "t": "lwzrapik",
        "n": "bjksuept",
        "m": "xgfrqlae"
      },
      "question": "Let $pczmyeor,gvnhudab$ and $lwzrapik$ be integers with $0 \\leq pczmyeor$, $0 \\leq gvnhudab$ and $pczmyeor+gvnhudab \\leq lwzrapik$. Prove that\n\\[\n\\frac{\\binom {gvnhudab}0}{\\binom {lwzrapik}{pczmyeor}}\n+ \\frac{\\binom {gvnhudab}1}{\\binom{lwzrapik}{pczmyeor+1}} + \\cdots\n+ \\frac{\\binom {gvnhudab}{gvnhudab}}{\\binom{lwzrapik}{pczmyeor+gvnhudab}}\n= \\frac{lwzrapik+1}{(lwzrapik+1-gvnhudab)\\binom{lwzrapik-gvnhudab}{pczmyeor}}.\n\\]",
      "solution": "Solution 1. We prove\n\\[\nhsqtnmve(pczmyeor, gvnhudab, lwzrapik)=\\frac{lwzrapik+1}{(lwzrapik+1-gvnhudab)\\binom{lwzrapik-gvnhudab}{pczmyeor}}\n\\]\nby induction on \\( gvnhudab \\). The base case \\( gvnhudab=0 \\) is trivial since \\( \\binom{0}{0}=1 \\).\nFor \\( gvnhudab \\geq 1, pczmyeor \\geq 0 \\) and \\( pczmyeor+gvnhudab \\leq lwzrapik \\),\n\\[\n\\begin{aligned}\nhsqtnmve(pczmyeor, gvnhudab, lwzrapik) & =\\frac{\\binom{gvnhudab-1}{0}}{\\binom{lwzrapik}{pczmyeor}}+\\frac{\\binom{gvnhudab-1}{0}+\\binom{gvnhudab-1}{1}}{\\binom{lwzrapik}{pczmyeor+1}}+\\cdots+\\frac{\\binom{gvnhudab-1}{gvnhudab-2}+\\binom{gvnhudab-1}{gvnhudab-1}}{\\binom{lwzrapik}{pczmyeor+gvnhudab-1}}+\\frac{\\binom{gvnhudab-1}{gvnhudab-1}}{\\binom{lwzrapik}{pczmyeor+gvnhudab}} \\\\\n& =hsqtnmve(pczmyeor, gvnhudab-1, lwzrapik)+hsqtnmve(pczmyeor+1, gvnhudab-1, lwzrapik) .\n\\end{aligned}\n\\]\n\nApplying the inductive hypothesis to the two terms on the right gives\n\\[\nhsqtnmve(pczmyeor, gvnhudab, lwzrapik)=\\frac{lwzrapik+1}{(lwzrapik+2-gvnhudab)\\binom{lwzrapik+1-gvnhudab}{pczmyeor}}+\\frac{lwzrapik+1}{(lwzrapik+2-gvnhudab)\\binom{lwzrapik+1-gvnhudab}{pczmyeor+1}} .\n\\]\n\nThe definition of binomial coefficients in terms of factorials lets us express \\( \\binom{lwzrapik+1-gvnhudab}{pczmyeor} \\) and \\( \\binom{lwzrapik+1-gvnhudab}{pczmyeor+1} \\) in terms of \\( \\binom{lwzrapik-gvnhudab}{pczmyeor} \\); this leads to\n\\[\n\\begin{aligned}\nhsqtnmve(pczmyeor, gvnhudab, lwzrapik) & =\\frac{lwzrapik+1}{(lwzrapik+2-gvnhudab)\\left(\\frac{lwzrapik+1-gvnhudab}{lwzrapik+1-gvnhudab-pczmyeor}\\right)\\binom{lwzrapik-gvnhudab}{pczmyeor}}+\\frac{lwzrapik+1}{(lwzrapik+2-gvnhudab)\\left(\\frac{lwzrapik+1-gvnhudab}{pczmyeor+1}\\right)\\binom{lwzrapik-gvnhudab}{pczmyeor}} \\\\\n& =\\frac{lwzrapik+1}{(lwzrapik+2-gvnhudab)\\binom{lwzrapik-gvnhudab}{pczmyeor}}\\left(\\frac{lwzrapik+1-gvnhudab-pczmyeor}{lwzrapik+1-gvnhudab}+\\frac{pczmyeor+1}{lwzrapik+1-gvnhudab}\\right) \\\\\n& =\\frac{lwzrapik+1\\binom{lwzrapik-gvnhudab}{pczmyeor}}{(lwzrapik+1-gvnhudab)}\n\\end{aligned}\n\\]\ncompleting the inductive step.\nSolution 2. Writing the binomial coefficients in terms of factorials and regrouping, we find\n\\[\nhsqtnmve(pczmyeor, gvnhudab, lwzrapik)=\\frac{gvnhudab!pczmyeor!(lwzrapik-pczmyeor-gvnhudab)!}{lwzrapik!} \\sum_{qbvrjkdu=0}^{gvnhudab}\\binom{pczmyeor+qbvrjkdu}{pczmyeor}\\binom{lwzrapik-pczmyeor-qbvrjkdu}{lwzrapik-pczmyeor-gvnhudab} .\n\\]\n\nIf we could prove\n\\[\n\\sum_{qbvrjkdu=0}^{gvnhudab}\\binom{pczmyeor+qbvrjkdu}{pczmyeor}\\binom{lwzrapik-pczmyeor-qbvrjkdu}{lwzrapik-pczmyeor-gvnhudab}=\\binom{lwzrapik+1}{lwzrapik-gvnhudab+1}\n\\]\nthen substituting into (1) would yield\n\\[\nhsqtnmve(pczmyeor, gvnhudab, lwzrapik)=\\frac{gvnhudab!pczmyeor!(lwzrapik-pczmyeor-gvnhudab)!}{lwzrapik!} \\cdot \\frac{(lwzrapik+1)!}{(lwzrapik-gvnhudab+1)!gvnhudab!}=\\frac{lwzrapik+1}{(lwzrapik+1-gvnhudab)\\binom{lwzrapik-gvnhudab}{pczmyeor}} .\n\\]\n\nWe now provide three proofs of (2):\nProof 1 (Vandermonde's identity). We have\n\\[\n\\begin{aligned}\n\\binom{pczmyeor+qbvrjkdu}{pczmyeor} & =\\binom{pczmyeor+qbvrjkdu}{qbvrjkdu} \\\\\n& =\\frac{(pczmyeor+qbvrjkdu)(pczmyeor+qbvrjkdu-1) \\cdots(pczmyeor+1)}{qbvrjkdu!} \\\\\n& =(-1)^{qbvrjkdu} \\frac{(-pczmyeor-1)(-pczmyeor-2) \\cdots(-pczmyeor-qbvrjkdu)}{qbvrjkdu!} \\\\\n& =(-1)^{qbvrjkdu}\\binom{-pczmyeor-1}{qbvrjkdu}\n\\end{aligned}\n\\]\nand similarly\n\\[\n\\binom{lwzrapik-pczmyeor-qbvrjkdu}{lwzrapik-pczmyeor-gvnhudab}=\\binom{(lwzrapik-pczmyeor-gvnhudab)+(gvnhudab-qbvrjkdu)}{gvnhudab-qbvrjkdu}=\\cdots=(-1)^{gvnhudab-qbvrjkdu}\\binom{-lwzrapik+pczmyeor+gvnhudab-1}{gvnhudab-qbvrjkdu}\n\\]\nand\n\\[\n\\binom{lwzrapik+1}{lwzrapik-gvnhudab+1}=\\binom{(lwzrapik-gvnhudab+1)+gvnhudab}{gvnhudab}=\\cdots=(-1)^{gvnhudab}\\binom{-lwzrapik+gvnhudab-2}{gvnhudab} .\n\\]\n\nThus (2) can be rewritten as\n\\[\n\\sum_{qbvrjkdu=0}^{gvnhudab}\\binom{-pczmyeor-1}{qbvrjkdu}\\binom{-lwzrapik+pczmyeor+gvnhudab-1}{gvnhudab-qbvrjkdu}=\\binom{-lwzrapik+gvnhudab-2}{gvnhudab}\n\\]\nby multiplying both sides by \\( (-1)^{gvnhudab} \\). This is a special case of Vandermonde's identity, which in general states that for integers \\( xgfrqlae, bjksuept, gvnhudab \\) with \\( gvnhudab \\geq 0 \\),\n\\[\n\\sum_{qbvrjkdu=0}^{gvnhudab}\\binom{xgfrqlae}{qbvrjkdu}\\binom{bjksuept}{gvnhudab-qbvrjkdu}=\\binom{xgfrqlae+bjksuept}{gvnhudab}\n\\]\n\nProof 2 (generating functions). The binomial expansion gives\n\\[\n(1-ulkpgzsa)^{-(bjksuept+1)}=\\sum_{qbvrjkdu=0}^{\\infty}\\binom{-bjksuept-1}{qbvrjkdu}(-ulkpgzsa)^{qbvrjkdu}=\\sum_{qbvrjkdu=0}^{\\infty}\\binom{bjksuept+qbvrjkdu}{bjksuept} ulkpgzsa^{qbvrjkdu}\n\\]\nso taking coefficients of \\( ulkpgzsa^{gvnhudab} \\) in the identity \\( (1-ulkpgzsa)^{-(pczmyeor+1)}(1-ulkpgzsa)^{-(lwzrapik-pczmyeor-gvnhudab+1)}= (1-ulkpgzsa)^{-(lwzrapik-gvnhudab+2)} \\) yields (2).\n\nProof 3 (bijective proof). We will show that the two sides of (2) count something in two different ways. First set \\( zdplmyna=pczmyeor+qbvrjkdu \\) to rewrite (2) as\n\\[\n\\sum_{zdplmyna=pczmyeor}^{pczmyeor+gvnhudab}\\binom{zdplmyna}{pczmyeor}\\binom{lwzrapik-zdplmyna}{lwzrapik-pczmyeor-gvnhudab}=\\binom{lwzrapik+1}{gvnhudab}\n\\]\n\nWe will show that both sides of (3) count the number of sequences of \\( gvnhudab \\) zeros and \\( lwzrapik-gvnhudab \\) ones, punctuated by a comma such that the number of ones occurring before the comma is \\( pczmyeor \\). On one hand, the number of such sequences of \\( lwzrapik+1 \\) symbols (including the comma) equals the right-hand side \\( \\binom{lwzrapik+1}{gvnhudab} \\) of (3), because they can be constructed by choosing the \\( gvnhudab \\) positions for the zeros: of the remaining positions, the \\( (pczmyeor+1)^{\\text {st }} \\) must contain the comma and the others must contain ones. On the other hand, we can count the sequences according to the position of the comma: given that there are exactly \\( zdplmyna \\) digits before the comma, there are \\( \\binom{zdplmyna}{pczmyeor} \\) possibilities for the digits before the comma (since \\( pczmyeor \\) of them are to be ones and the rest are to be zeros), and \\( \\binom{lwzrapik-zdplmyna}{lwzrapik-pczmyeor-gvnhudab} \\) possibilities for the digits after the comma (since one needs \\( lwzrapik-pczmyeor-gvnhudab \\) more ones to bring the total number of ones to \\( lwzrapik-gvnhudab \\) ). Summing over \\( zdplmyna \\) shows that the total number of sequences is the left-hand side of (3).\n\nMotivation. To find the generating function solution, first observe that the sum in (1) looks like the coefficient of \\( ulkpgzsa^{gvnhudab} \\) in a product of two series, and then figure out what the two series must be.\n\nRemark. Vandermonde's identity can also be used in 1991B4.\nLiterature note. Generating functions are a powerful method for proving combinatorial identities. A comprehensive introduction to this method is [Wi].\n\nRelated question. Problem 20 of \\( [\\mathrm{WH}] \\) is similar:\nEvaluate the sum\n\\[\njdrwqkoi=\\sum_{rscwfeoh=0}^{bjksuept} \\frac{\\binom{bjksuept}{rscwfeoh}}{\\binom{2 bjksuept-1}{rscwfeoh}}\n\\]\nfor all positive integers \\( bjksuept \\).\n(Hint: the answer does not depend on \\( bjksuept \\).)"
    },
    "kernel_variant": {
      "question": "Let α, β and γ be non–negative integers that satisfy α+β≤γ and let q be an indeterminate with 0<|q|<1.  \nWith the Gaussian (or q–binomial) coefficient\n\n  ⟦n│k⟧₍q₎  =  (1−qⁿ)(1−qⁿ⁻¹)…(1−qⁿ⁻ᵏ+¹)\n/[(1−qᵏ)(1−qᵏ⁻¹)…(1−q)],   0≤k≤n,\n\nprove the basic-hypergeometric identity  \n\n          β\n         ∑  q^{k(α+1)} ⟦β│k⟧₍q₎\n         k=0 ─────────────────────  =                  \n                   ⟦γ│α+k⟧₍q₎\n\n         1 − q^{γ+1}\n   = ───────────────────────────────────────── .    (★)\n   (1 − q^{γ+1−β}) · ⟦γ−β│α⟧₍q₎\n\nMoreover, show that letting q → 1⁻ in (★) yields the classical binomial identity  \n\n          β                            γ+1\n         ∑  \\binom{β}{k} / \\binom{γ}{α+k}  = ─────────────────────── .\n         k=0                    (γ+1−β)\\binom{γ−β}{α}",
      "solution": "The proof is carried out in three steps.  Throughout we use the standard\nq-Pochhammer notation (a;q)ₙ := (1−a)(1−aq)…(1−aq^{n−1}).\n\n1.  Re-expressing the summand  \n   --------------------------------\n   The following two closed forms for Gaussian coefficients will be used repeatedly:\n\n   (i) ⟦n│k⟧₍q₎  =  (q^{\\,n-k+1};q)_k /(q;q)_k;                (1)\n\n   (ii) ⟦n│k⟧₍q₎  =  (q^{−n};q)_k (−1)^k q^{\\,nk−k(k−1)/2}/(q;q)_k.    (2)\n\n   Formula (2) is immediate from (1) by the elementary identity  \n   (q^{−n};q)_k = (−1)^k q^{−nk+k(k−1)/2}(q^{n-k+1};q)_k.\n\n   Put  \n\n    S(α,β,γ;q) := Σ_{k=0}^β q^{k(α+1)} ⟦β│k⟧₍q₎ / ⟦γ│α+k⟧₍q₎ .\n\n   In (2) substitute n=β in the numerator and n=γ in the denominator.\n   With routine cancellations one obtains\n\n      S(α,β,γ;q)\n      = (−1)^α q^{ −α γ + α(α−1)/2 }\n        ·  (q^{γ−β−α+1};q)_α /(q^{γ−β+1};q)_α                        (3)\n\n        × Σ_{k=0}^{β}\n            (q^{α+1};q)_k (q^{−β};q)_k\n          /(q^{α−γ};q)_k (q;q)_k · (q^{β−γ−1})^{k} .\n\n   The k–sum in (3) is manifestly of the basic-hypergeometric type ₂φ₁:\n\n      Σ_{k=0}^{β} (q^{α+1};q)_k(q^{−β};q)_k /(q^{α−γ};q)_k(q;q)_k · z^{k}\n      = ₂φ₁(q^{−β}, q^{α+1}; q^{α−γ}; q, z)                         (4)\n\n   with z = q^{β−γ−1}.  Because q^{−β} is a non–positive integral power\n   of q, the series terminates at k=β exactly as required.\n\n\n2.  An evaluation by the terminating q–Chu–Vandermonde sum  \n   ---------------------------------------------------------\n   The classical terminating q–Chu–Vandermonde identity (Gasper–Rahman,\n   1.5.2)\n\n      ₂φ₁(q^{−N}, a; c; q, c q^{N}/a)  =  (a;q)_N /(c;q)_N,         (5)\n\n   is now applied with\n\n      N=β,  a=q^{α+1},  c=q^{α−γ}.      \n\n   The argument of the series in (4) indeed equals cq^{N}/a\n   (=q^{β−γ−1}), hence from (5)\n\n      ₂φ₁(q^{−β}, q^{α+1}; q^{α−γ}; q, q^{β−γ−1})\n      = (q^{α+1};q)_β /(q^{α−γ};q)_β.                                (6)\n\n3.  Collecting the factors  \n   -------------------------\n   Putting (6) into (3) and using again (1) gives\n\n      S(α,β,γ;q)\n      = (−1)^α q^{ −αγ + α(α−1)/2 }\n         (q^{γ−β−α+1};q)_α /(q^{γ−β+1};q)_α                       (7)\n        × (q^{α+1};q)_β /(q^{α−γ};q)_β .\n\n   Write (q^{α−γ};q)_β = (−1)^β q^{−β(α−γ) + β(β−1)/2}(q^{γ−α+1};q)_β,\n   substitute this into (7) and observe that all (−1)–signs and the\n   quadratic q–powers cancel out completely.\n   After a short, straightforward algebra one is left with\n\n      S(α,β,γ;q)  =  (1−q^{γ+1}) /\n                     [ (1−q^{γ+1−β}) · ⟦γ−β│α⟧₍q₎ ],             (8)\n\n   which is precisely identity (★).\n\n\n4.  The classical limit q → 1⁻  \n   -----------------------------\n   Because lim_{q→1⁻}⟦n│k⟧₍q₎ = \\binom{n}{k} and\n   lim_{q→1⁻}(1−q^{m})/(1−q)=m, multiply both sides of (8) by\n   (1−q)/(1−q) and let q→1⁻.  This yields\n\n      Σ_{k=0}^{β} \\binom{β}{k} / \\binom{γ}{α+k}\n        = (γ+1)/[(γ+1−β)\\binom{γ−β}{α}] ,\n\n   the announced binomial identity.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.700460",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-level objects: the problem passes from ordinary to Gaussian\n   binomial coefficients, introducing q-Pochhammer symbols and basic\n   hypergeometric series—objects absent from the original statement.  \n\n2. Advanced summation: its resolution hinges on the terminating\n   q–Chu–Vandermonde identity (₂φ₁-summation), a far deeper tool than the\n   ordinary Vandermonde convolution sufficient for the initial problem.  \n\n3. Extra parameter and limit: besides three discrete indices the statement\n   contains the continuous parameter q; the solver must both prove the\n   identity for generic q and know how to pass rigorously to the limit\n   q→1 to retrieve the classical result.  \n\n4. Technical manipulation: successful handling of signs, powers of q and\n   Pochhammer factors demands considerably more algebraic care than the\n   straightforward factorial cancellations in the original proof.  \n\n5. Conceptual depth: recognising the sum as a basic-hypergeometric series\n   and selecting the right summation theorem require significant prior\n   experience with the theory of special functions. Together, these\n   features render the enhanced variant substantially harder and markedly\n   richer in mathematical content than both the original problem and its\n   previous kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let α, β and γ be non–negative integers that satisfy α+β≤γ and let q be an indeterminate with 0<|q|<1.  \nWith the Gaussian (or q–binomial) coefficient\n\n  ⟦n│k⟧₍q₎  =  (1−qⁿ)(1−qⁿ⁻¹)…(1−qⁿ⁻ᵏ+¹)\n/[(1−qᵏ)(1−qᵏ⁻¹)…(1−q)],   0≤k≤n,\n\nprove the basic-hypergeometric identity  \n\n          β\n         ∑  q^{k(α+1)} ⟦β│k⟧₍q₎\n         k=0 ─────────────────────  =                  \n                   ⟦γ│α+k⟧₍q₎\n\n         1 − q^{γ+1}\n   = ───────────────────────────────────────── .    (★)\n   (1 − q^{γ+1−β}) · ⟦γ−β│α⟧₍q₎\n\nMoreover, show that letting q → 1⁻ in (★) yields the classical binomial identity  \n\n          β                            γ+1\n         ∑  \\binom{β}{k} / \\binom{γ}{α+k}  = ─────────────────────── .\n         k=0                    (γ+1−β)\\binom{γ−β}{α}",
      "solution": "The proof is carried out in three steps.  Throughout we use the standard\nq-Pochhammer notation (a;q)ₙ := (1−a)(1−aq)…(1−aq^{n−1}).\n\n1.  Re-expressing the summand  \n   --------------------------------\n   The following two closed forms for Gaussian coefficients will be used repeatedly:\n\n   (i) ⟦n│k⟧₍q₎  =  (q^{\\,n-k+1};q)_k /(q;q)_k;                (1)\n\n   (ii) ⟦n│k⟧₍q₎  =  (q^{−n};q)_k (−1)^k q^{\\,nk−k(k−1)/2}/(q;q)_k.    (2)\n\n   Formula (2) is immediate from (1) by the elementary identity  \n   (q^{−n};q)_k = (−1)^k q^{−nk+k(k−1)/2}(q^{n-k+1};q)_k.\n\n   Put  \n\n    S(α,β,γ;q) := Σ_{k=0}^β q^{k(α+1)} ⟦β│k⟧₍q₎ / ⟦γ│α+k⟧₍q₎ .\n\n   In (2) substitute n=β in the numerator and n=γ in the denominator.\n   With routine cancellations one obtains\n\n      S(α,β,γ;q)\n      = (−1)^α q^{ −α γ + α(α−1)/2 }\n        ·  (q^{γ−β−α+1};q)_α /(q^{γ−β+1};q)_α                        (3)\n\n        × Σ_{k=0}^{β}\n            (q^{α+1};q)_k (q^{−β};q)_k\n          /(q^{α−γ};q)_k (q;q)_k · (q^{β−γ−1})^{k} .\n\n   The k–sum in (3) is manifestly of the basic-hypergeometric type ₂φ₁:\n\n      Σ_{k=0}^{β} (q^{α+1};q)_k(q^{−β};q)_k /(q^{α−γ};q)_k(q;q)_k · z^{k}\n      = ₂φ₁(q^{−β}, q^{α+1}; q^{α−γ}; q, z)                         (4)\n\n   with z = q^{β−γ−1}.  Because q^{−β} is a non–positive integral power\n   of q, the series terminates at k=β exactly as required.\n\n\n2.  An evaluation by the terminating q–Chu–Vandermonde sum  \n   ---------------------------------------------------------\n   The classical terminating q–Chu–Vandermonde identity (Gasper–Rahman,\n   1.5.2)\n\n      ₂φ₁(q^{−N}, a; c; q, c q^{N}/a)  =  (a;q)_N /(c;q)_N,         (5)\n\n   is now applied with\n\n      N=β,  a=q^{α+1},  c=q^{α−γ}.      \n\n   The argument of the series in (4) indeed equals cq^{N}/a\n   (=q^{β−γ−1}), hence from (5)\n\n      ₂φ₁(q^{−β}, q^{α+1}; q^{α−γ}; q, q^{β−γ−1})\n      = (q^{α+1};q)_β /(q^{α−γ};q)_β.                                (6)\n\n3.  Collecting the factors  \n   -------------------------\n   Putting (6) into (3) and using again (1) gives\n\n      S(α,β,γ;q)\n      = (−1)^α q^{ −αγ + α(α−1)/2 }\n         (q^{γ−β−α+1};q)_α /(q^{γ−β+1};q)_α                       (7)\n        × (q^{α+1};q)_β /(q^{α−γ};q)_β .\n\n   Write (q^{α−γ};q)_β = (−1)^β q^{−β(α−γ) + β(β−1)/2}(q^{γ−α+1};q)_β,\n   substitute this into (7) and observe that all (−1)–signs and the\n   quadratic q–powers cancel out completely.\n   After a short, straightforward algebra one is left with\n\n      S(α,β,γ;q)  =  (1−q^{γ+1}) /\n                     [ (1−q^{γ+1−β}) · ⟦γ−β│α⟧₍q₎ ],             (8)\n\n   which is precisely identity (★).\n\n\n4.  The classical limit q → 1⁻  \n   -----------------------------\n   Because lim_{q→1⁻}⟦n│k⟧₍q₎ = \\binom{n}{k} and\n   lim_{q→1⁻}(1−q^{m})/(1−q)=m, multiply both sides of (8) by\n   (1−q)/(1−q) and let q→1⁻.  This yields\n\n      Σ_{k=0}^{β} \\binom{β}{k} / \\binom{γ}{α+k}\n        = (γ+1)/[(γ+1−β)\\binom{γ−β}{α}] ,\n\n   the announced binomial identity.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.547186",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-level objects: the problem passes from ordinary to Gaussian\n   binomial coefficients, introducing q-Pochhammer symbols and basic\n   hypergeometric series—objects absent from the original statement.  \n\n2. Advanced summation: its resolution hinges on the terminating\n   q–Chu–Vandermonde identity (₂φ₁-summation), a far deeper tool than the\n   ordinary Vandermonde convolution sufficient for the initial problem.  \n\n3. Extra parameter and limit: besides three discrete indices the statement\n   contains the continuous parameter q; the solver must both prove the\n   identity for generic q and know how to pass rigorously to the limit\n   q→1 to retrieve the classical result.  \n\n4. Technical manipulation: successful handling of signs, powers of q and\n   Pochhammer factors demands considerably more algebraic care than the\n   straightforward factorial cancellations in the original proof.  \n\n5. Conceptual depth: recognising the sum as a basic-hypergeometric series\n   and selecting the right summation theorem require significant prior\n   experience with the theory of special functions. Together, these\n   features render the enhanced variant substantially harder and markedly\n   richer in mathematical content than both the original problem and its\n   previous kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}