Dataset Card for numina-deepseek-r1-qwen-14b

This dataset has been created with distilabel.

Dataset Summary

This dataset contains a pipeline.yaml which can be used to reproduce the pipeline that generated it in distilabel using the distilabel CLI:

distilabel pipeline run --config "https://huggingface.co/datasets/mikehps123/numina-deepseek-r1-qwen-14b/raw/main/pipeline.yaml"

or explore the configuration:

distilabel pipeline info --config "https://huggingface.co/datasets/mikehps123/numina-deepseek-r1-qwen-14b/raw/main/pipeline.yaml"

Dataset structure

The examples have the following structure per configuration:

Configuration: default

---

{
    "distilabel_metadata": {
        "raw_input_text_generation_0": [
            {
                "content": "You will be given a problem. Please reason step by step, and put your final answer within \boxed{}:\nWhat is the coefficient of $x^2y^6$ in the expansion of $\\left(\\frac{3}{5}x-\\frac{y}{2}\\right)^8$?  Express your answer as a common fraction.",
                "role": "user"
            }
        ],
        "raw_output_text_generation_0": "Alright, so I\u0027ve got this problem here: I need to find the coefficient of $x^2{y}^{6}x^2\; y^6$ in the expansion of $left(frac35x-fracy2right{)}^8\backslash \backslash left(\backslash \backslash frac\{3\}\{5\}x\; -\; \backslash \backslash frac\{y\}\{2\}\backslash \backslash right)^8$. Hmm, okay. Let me think about how to approach this.\n\nFirst, I remember that when you have a binomial expression raised to a power, you can use the binomial theorem to expand it. The binomial theorem states that $(a+b{)}^n=su{m}_{k=0}^{n}binomnk{a}^{n-k}{b}^k(a\; +\; b)^n\; =\; \backslash \backslash sum\_\{k=0\}^\{n\}\; \backslash \backslash binom\{n\}\{k\}\; a^\{n\; -\; k\}\; b^k$. So in this case, my $aa$ is $frac35x\backslash \backslash frac\{3\}\{5\}x$ and my $bb$ is $-fracy2-\backslash \backslash frac\{y\}\{2\}$, right? And $nn$ is 8.\n\nI need to find the term where the power of $xx$ is 2 and the power of $yy$ is 6. That means I\u0027m looking for the term where $aa$ is raised to the 2nd power and $bb$ is raised to the 6th power. \n\nSo, using the binomial theorem, the general term is $binom8kleft(frac35xright{)}^{8-k}left(-fracy2right{)}^k\backslash \backslash binom\{8\}\{k\}\; \backslash \backslash left(\backslash \backslash frac\{3\}\{5\}x\backslash \backslash right)^\{8\; -\; k\}\; \backslash \backslash left(-\backslash \backslash frac\{y\}\{2\}\backslash \backslash right)^k$. I need to find the value of $kk$ such that the exponents on $xx$ and $yy$ match what I\u0027m looking for. \n\nSince $xx$ is in the term $left(frac35xright{)}^{8-k}\backslash \backslash left(\backslash \backslash frac\{3\}\{5\}x\backslash \backslash right)^\{8\; -\; k\}$, the exponent on $xx$ is $8-k8\; -\; k$. Similarly, the term $left(-fracy2right{)}^k\backslash \backslash left(-\backslash \backslash frac\{y\}\{2\}\backslash \backslash right)^k$ will have an exponent of $kk$ on $yy$. \n\nI want the exponent on $xx$ to be 2, so $8-k=28\; -\; k\; =\; 2$. Solving for $kk$, I get $k=8-2=6k\; =\; 8\; -\; 2\; =\; 6$. Perfect, that means the exponent on $yy$ will be 6 as required. So the term I need is when $k=6k\; =\; 6$.\n\nNow, let\u0027s write out that specific term:\n\n\\[\n\\binom{8}{6} \\left(\\frac{3}{5}x\\right)^{2} \\left(-\\frac{y}{2}\\right)^6\n\\]\n\nOkay, let\u0027s compute each part step by step.\n\nFirst, the binomial coefficient $binom86\backslash \backslash binom\{8\}\{6\}$. Remember that $binomnk=fracn!k!(n-k)!\backslash \backslash binom\{n\}\{k\}\; =\; \backslash \backslash frac\{n!\}\{k!(n\; -\; k)!\}$. So plugging in the numbers:\n\n\\[\n\\binom{8}{6} = \\frac{8!}{6! \\cdot (8 - 6)!} = \\frac{8!}{6! \\cdot 2!}\n\\]\n\nCalculating factorial values: $8!=403208!\; =\; 40320$, $6!=7206!\; =\; 720$, and $2!=22!\; =\; 2$. So:\n\n\\[\n\\frac{40320}{720 \\cdot 2} = \\frac{40320}{1440} = 28\n\\]\n\nAlright, so the binomial coefficient is 28.\n\nNext, let\u0027s compute $left(frac35xright{)}^2\backslash \backslash left(\backslash \backslash frac\{3\}\{5\}x\backslash \backslash right)^2$. That\u0027s $left(frac35right{)}^2{x}^2\backslash \backslash left(\backslash \backslash frac\{3\}\{5\}\backslash \backslash right)^2\; x^2$. Calculating $left(frac35right{)}^2\backslash \backslash left(\backslash \backslash frac\{3\}\{5\}\backslash \backslash right)^2$:\n\n\\[\n\\frac{3^2}{5^2} = \\frac{9}{25}\n\\]\n\nSo that part is $frac925x^2\backslash \backslash frac\{9\}\{25\}x^2$.\n\nNow, the last part is $left(-fracy2right{)}^6\backslash \backslash left(-\backslash \backslash frac\{y\}\{2\}\backslash \backslash right)^6$. Let\u0027s calculate that. First, the coefficient: $left(-frac12right{)}^6\backslash \backslash left(-\backslash \backslash frac\{1\}\{2\}\backslash \backslash right)^6$. Since the exponent is even, the negative sign will become positive.\n\n\\[\n\\left(-\\frac{1}{2}\\right)^6 = \\left(\\frac{1}{2}\\right)^6 = \\frac{1}{64}\n\\]\n\nSo, that part is $frac164y^6\backslash \backslash frac\{1\}\{64\}y^6$.\n\nPutting it all together, the term is:\n\n\\[\n28 \\cdot \\frac{9}{25}x^2 \\cdot \\frac{1}{64}y^6\n\\]\n\nNow, let\u0027s compute the coefficients:\n\nFirst, multiply the numerical coefficients: 28, 9/25, and 1/64.\n\nLet me write them all as fractions for ease:\n\n28 is $frac281\backslash \backslash frac\{28\}\{1\}$, 9/25 stays the same, and 1/64 stays the same.\n\nSo, multiplying:\n\n\\[\n\\frac{28}{1} \\times \\frac{9}{25} \\times \\frac{1}{64}\n\\]\n\nMultiply the numerators: 28 * 9 * 1 = 252\n\nMultiply the denominators: 1 * 25 * 64 = 1600\n\nSo, the coefficient is $frac2521600\backslash \backslash frac\{252\}\{1600\}$.\n\nNow, let\u0027s simplify that fraction. Both numerator and denominator are divisible by 4.\n\n252 \u00f7 4 = 63\n\n1600 \u00f7 4 = 400\n\nSo, simplifying, we get $frac63400\backslash \backslash frac\{63\}\{400\}$.\n\nWait, is that reducible further? Let\u0027s see. 63 and 400.\n\n63 factors: 7 * 9 = 7 * 3^2\n\n400 factors: 16 * 25 = 2^4 * 5^2\n\nNo common factors, so $frac63400\backslash \backslash frac\{63\}\{400\}$ is the simplified form.\n\nTherefore, the coefficient of $x^2{y}^{6}x^2\; y^6$ in the expansion is $frac63400\backslash \backslash frac\{63\}\{400\}$.\n\nWait a second, just double-checking my work. Let me go through the calculations again to make sure I didn\u0027t make a mistake.\n\nFirst, binomial coefficient: $binom86=28\backslash \backslash binom\{8\}\{6\}\; =\; 28$. That seems right because $binom82\backslash \backslash binom\{8\}\{2\}$ is also 28, and since $binomnk=binomnn-k\backslash \backslash binom\{n\}\{k\}\; =\; \backslash \backslash binom\{n\}\{n\; -\; k\}$, that\u0027s correct.\n\nNext, $left(frac35right{)}^2=frac925\backslash \backslash left(\backslash \backslash frac\{3\}\{5\}\backslash \backslash right)^2\; =\; \backslash \backslash frac\{9\}\{25\}$. Correct.\n\nThen, $left(-frac12right{)}^6=frac164\backslash \backslash left(-\backslash \backslash frac\{1\}\{2\}\backslash \backslash right)^6\; =\; \backslash \backslash frac\{1\}\{64\}$. Correct, because the exponent is even, so the negative becomes positive.\n\nThen, multiplying 28 * 9 * 1 = 252. Correct.\n\nMultiplying denominators: 25 * 64 = 1600. Wait, 25 * 64 is 1600? Let me check: 25 * 60 is 1500, and 25 * 4 is 100, so yes, 1500 + 100 = 1600. Correct.\n\nSimplify $frac2521600\backslash \backslash frac\{252\}\{1600\}$. Divided by 4: 252 \u00f7 4 = 63, 1600 \u00f7 4 = 400. Correct.\n\nAnd 63 and 400 have no common factors besides 1. So, yeah, $frac63400\backslash \backslash frac\{63\}\{400\}$ is the simplified fraction.\n\nSo, I think I did that correctly.\n\nJust to recap, the term is $binom86timesleft(frac35right{)}^2{x}^{2}timesleft(-frac12right{)}^6{y}^6\backslash \backslash binom\{8\}\{6\}\; \backslash \backslash times\; \backslash \backslash left(\backslash \backslash frac\{3\}\{5\}\backslash \backslash right)^2\; x^2\; \backslash \backslash times\; \backslash \backslash left(-\backslash \backslash frac\{1\}\{2\}\backslash \backslash right)^6\; y^6$. Calculated each part, multiplied them together, and simplified the fraction.\n\nYep, that seems solid. So, the coefficient is $frac63400\backslash \backslash frac\{63\}\{400\}$.\n\n**Final Answer**\nThe coefficient is \\boxed{\\dfrac{63}{400}}.\n\u003c/think\u003e\n\nTo find the coefficient of $x^2{y}^{6}x^2\; y^6$ in the expansion of $left(frac35x-fracy2right{)}^8\backslash \backslash left(\backslash \backslash frac\{3\}\{5\}x\; -\; \backslash \backslash frac\{y\}\{2\}\backslash \backslash right)^8$, we use the binomial theorem. The binomial theorem states that $(a+b{)}^n=su{m}_{k=0}^{n}binomnk{a}^{n-k}{b}^k(a\; +\; b)^n\; =\; \backslash \backslash sum\_\{k=0\}^\{n\}\; \backslash \backslash binom\{n\}\{k\}\; a^\{n\; -\; k\}\; b^k$. Here, $a=frac35xa\; =\; \backslash \backslash frac\{3\}\{5\}x$, $b=-fracy2b\; =\; -\backslash \backslash frac\{y\}\{2\}$, and $n=8n\; =\; 8$.\n\nWe need the term where the exponents of $xx$ and $yy$ are 2 and 6, respectively. This corresponds to $k=6k\; =\; 6$ because the exponent on $xx$ is $8-k=28\; -\; k\; =\; 2$ and the exponent on $yy$ is $k=6k\; =\; 6$.\n\nThe specific term is:\n\\[\n\\binom{8}{6} \\left(\\frac{3}{5}x\\right)^2 \\left(-\\frac{y}{2}\\right)^6\n\\]\n\nFirst, compute the binomial coefficient:\n\\[\n\\binom{8}{6} = \\frac{8!}{6! \\cdot 2!} = \\frac{40320}{720 \\cdot 2} = 28\n\\]\n\nNext, compute $left(frac35right{)}^2\backslash \backslash left(\backslash \backslash frac\{3\}\{5\}\backslash \backslash right)^2$:\n\\[\n\\left(\\frac{3}{5}\\right)^2 = \\frac{9}{25}\n\\]\n\nThen, compute $left(-frac12right{)}^6\backslash \backslash left(-\backslash \backslash frac\{1\}\{2\}\backslash \backslash right)^6$:\n\\[\n\\left(-\\frac{1}{2}\\right)^6 = \\frac{1}{64}\n\\]\n\nPutting it all together, the term is:\n\\[\n28 \\cdot \\frac{9}{25} \\cdot \\frac{1}{64}\n\\]\n\nMultiply the numerical coefficients:\n\\[\n28 \\times \\frac{9}{25} \\times \\frac{1}{64} = \\frac{28 \\times 9 \\times 1}{25 \\times 64} = \\frac{252}{1600}\n\\]\n\nSimplify the fraction:\n\\[\n\\frac{252}{1600} = \\frac{63}{400}\n\\]\n\nThus, the coefficient of $x^2{y}^{6}x^2\; y^6$ is $boxeddfrac63400\backslash \backslash boxed\{\backslash \backslash dfrac\{63\}\{400\}\}$.",
        "statistics_text_generation_0": {
            "input_tokens": 79,
            "output_tokens": 2361
        }
    },
    "generation": "Alright, so I\u0027ve got this problem here: I need to find the coefficient of $x^2{y}^{6}x^2\; y^6$ in the expansion of $left(frac35x-fracy2right{)}^8\backslash \backslash left(\backslash \backslash frac\{3\}\{5\}x\; -\; \backslash \backslash frac\{y\}\{2\}\backslash \backslash right)^8$. Hmm, okay. Let me think about how to approach this.\n\nFirst, I remember that when you have a binomial expression raised to a power, you can use the binomial theorem to expand it. The binomial theorem states that $(a+b{)}^n=su{m}_{k=0}^{n}binomnk{a}^{n-k}{b}^k(a\; +\; b)^n\; =\; \backslash \backslash sum\_\{k=0\}^\{n\}\; \backslash \backslash binom\{n\}\{k\}\; a^\{n\; -\; k\}\; b^k$. So in this case, my $aa$ is $frac35x\backslash \backslash frac\{3\}\{5\}x$ and my $bb$ is $-fracy2-\backslash \backslash frac\{y\}\{2\}$, right? And $nn$ is 8.\n\nI need to find the term where the power of $xx$ is 2 and the power of $yy$ is 6. That means I\u0027m looking for the term where $aa$ is raised to the 2nd power and $bb$ is raised to the 6th power. \n\nSo, using the binomial theorem, the general term is $binom8kleft(frac35xright{)}^{8-k}left(-fracy2right{)}^k\backslash \backslash binom\{8\}\{k\}\; \backslash \backslash left(\backslash \backslash frac\{3\}\{5\}x\backslash \backslash right)^\{8\; -\; k\}\; \backslash \backslash left(-\backslash \backslash frac\{y\}\{2\}\backslash \backslash right)^k$. I need to find the value of $kk$ such that the exponents on $xx$ and $yy$ match what I\u0027m looking for. \n\nSince $xx$ is in the term $left(frac35xright{)}^{8-k}\backslash \backslash left(\backslash \backslash frac\{3\}\{5\}x\backslash \backslash right)^\{8\; -\; k\}$, the exponent on $xx$ is $8-k8\; -\; k$. Similarly, the term $left(-fracy2right{)}^k\backslash \backslash left(-\backslash \backslash frac\{y\}\{2\}\backslash \backslash right)^k$ will have an exponent of $kk$ on $yy$. \n\nI want the exponent on $xx$ to be 2, so $8-k=28\; -\; k\; =\; 2$. Solving for $kk$, I get $k=8-2=6k\; =\; 8\; -\; 2\; =\; 6$. Perfect, that means the exponent on $yy$ will be 6 as required. So the term I need is when $k=6k\; =\; 6$.\n\nNow, let\u0027s write out that specific term:\n\n\\[\n\\binom{8}{6} \\left(\\frac{3}{5}x\\right)^{2} \\left(-\\frac{y}{2}\\right)^6\n\\]\n\nOkay, let\u0027s compute each part step by step.\n\nFirst, the binomial coefficient $binom86\backslash \backslash binom\{8\}\{6\}$. Remember that $binomnk=fracn!k!(n-k)!\backslash \backslash binom\{n\}\{k\}\; =\; \backslash \backslash frac\{n!\}\{k!(n\; -\; k)!\}$. So plugging in the numbers:\n\n\\[\n\\binom{8}{6} = \\frac{8!}{6! \\cdot (8 - 6)!} = \\frac{8!}{6! \\cdot 2!}\n\\]\n\nCalculating factorial values: $8!=403208!\; =\; 40320$, $6!=7206!\; =\; 720$, and $2!=22!\; =\; 2$. So:\n\n\\[\n\\frac{40320}{720 \\cdot 2} = \\frac{40320}{1440} = 28\n\\]\n\nAlright, so the binomial coefficient is 28.\n\nNext, let\u0027s compute $left(frac35xright{)}^2\backslash \backslash left(\backslash \backslash frac\{3\}\{5\}x\backslash \backslash right)^2$. That\u0027s $left(frac35right{)}^2{x}^2\backslash \backslash left(\backslash \backslash frac\{3\}\{5\}\backslash \backslash right)^2\; x^2$. Calculating $left(frac35right{)}^2\backslash \backslash left(\backslash \backslash frac\{3\}\{5\}\backslash \backslash right)^2$:\n\n\\[\n\\frac{3^2}{5^2} = \\frac{9}{25}\n\\]\n\nSo that part is $frac925x^2\backslash \backslash frac\{9\}\{25\}x^2$.\n\nNow, the last part is $left(-fracy2right{)}^6\backslash \backslash left(-\backslash \backslash frac\{y\}\{2\}\backslash \backslash right)^6$. Let\u0027s calculate that. First, the coefficient: $left(-frac12right{)}^6\backslash \backslash left(-\backslash \backslash frac\{1\}\{2\}\backslash \backslash right)^6$. Since the exponent is even, the negative sign will become positive.\n\n\\[\n\\left(-\\frac{1}{2}\\right)^6 = \\left(\\frac{1}{2}\\right)^6 = \\frac{1}{64}\n\\]\n\nSo, that part is $frac164y^6\backslash \backslash frac\{1\}\{64\}y^6$.\n\nPutting it all together, the term is:\n\n\\[\n28 \\cdot \\frac{9}{25}x^2 \\cdot \\frac{1}{64}y^6\n\\]\n\nNow, let\u0027s compute the coefficients:\n\nFirst, multiply the numerical coefficients: 28, 9/25, and 1/64.\n\nLet me write them all as fractions for ease:\n\n28 is $frac281\backslash \backslash frac\{28\}\{1\}$, 9/25 stays the same, and 1/64 stays the same.\n\nSo, multiplying:\n\n\\[\n\\frac{28}{1} \\times \\frac{9}{25} \\times \\frac{1}{64}\n\\]\n\nMultiply the numerators: 28 * 9 * 1 = 252\n\nMultiply the denominators: 1 * 25 * 64 = 1600\n\nSo, the coefficient is $frac2521600\backslash \backslash frac\{252\}\{1600\}$.\n\nNow, let\u0027s simplify that fraction. Both numerator and denominator are divisible by 4.\n\n252 \u00f7 4 = 63\n\n1600 \u00f7 4 = 400\n\nSo, simplifying, we get $frac63400\backslash \backslash frac\{63\}\{400\}$.\n\nWait, is that reducible further? Let\u0027s see. 63 and 400.\n\n63 factors: 7 * 9 = 7 * 3^2\n\n400 factors: 16 * 25 = 2^4 * 5^2\n\nNo common factors, so $frac63400\backslash \backslash frac\{63\}\{400\}$ is the simplified form.\n\nTherefore, the coefficient of $x^2{y}^{6}x^2\; y^6$ in the expansion is $frac63400\backslash \backslash frac\{63\}\{400\}$.\n\nWait a second, just double-checking my work. Let me go through the calculations again to make sure I didn\u0027t make a mistake.\n\nFirst, binomial coefficient: $binom86=28\backslash \backslash binom\{8\}\{6\}\; =\; 28$. That seems right because $binom82\backslash \backslash binom\{8\}\{2\}$ is also 28, and since $binomnk=binomnn-k\backslash \backslash binom\{n\}\{k\}\; =\; \backslash \backslash binom\{n\}\{n\; -\; k\}$, that\u0027s correct.\n\nNext, $left(frac35right{)}^2=frac925\backslash \backslash left(\backslash \backslash frac\{3\}\{5\}\backslash \backslash right)^2\; =\; \backslash \backslash frac\{9\}\{25\}$. Correct.\n\nThen, $left(-frac12right{)}^6=frac164\backslash \backslash left(-\backslash \backslash frac\{1\}\{2\}\backslash \backslash right)^6\; =\; \backslash \backslash frac\{1\}\{64\}$. Correct, because the exponent is even, so the negative becomes positive.\n\nThen, multiplying 28 * 9 * 1 = 252. Correct.\n\nMultiplying denominators: 25 * 64 = 1600. Wait, 25 * 64 is 1600? Let me check: 25 * 60 is 1500, and 25 * 4 is 100, so yes, 1500 + 100 = 1600. Correct.\n\nSimplify $frac2521600\backslash \backslash frac\{252\}\{1600\}$. Divided by 4: 252 \u00f7 4 = 63, 1600 \u00f7 4 = 400. Correct.\n\nAnd 63 and 400 have no common factors besides 1. So, yeah, $frac63400\backslash \backslash frac\{63\}\{400\}$ is the simplified fraction.\n\nSo, I think I did that correctly.\n\nJust to recap, the term is $binom86timesleft(frac35right{)}^2{x}^{2}timesleft(-frac12right{)}^6{y}^6\backslash \backslash binom\{8\}\{6\}\; \backslash \backslash times\; \backslash \backslash left(\backslash \backslash frac\{3\}\{5\}\backslash \backslash right)^2\; x^2\; \backslash \backslash times\; \backslash \backslash left(-\backslash \backslash frac\{1\}\{2\}\backslash \backslash right)^6\; y^6$. Calculated each part, multiplied them together, and simplified the fraction.\n\nYep, that seems solid. So, the coefficient is $frac63400\backslash \backslash frac\{63\}\{400\}$.\n\n**Final Answer**\nThe coefficient is \\boxed{\\dfrac{63}{400}}.\n\u003c/think\u003e\n\nTo find the coefficient of $x^2{y}^{6}x^2\; y^6$ in the expansion of $left(frac35x-fracy2right{)}^8\backslash \backslash left(\backslash \backslash frac\{3\}\{5\}x\; -\; \backslash \backslash frac\{y\}\{2\}\backslash \backslash right)^8$, we use the binomial theorem. The binomial theorem states that $(a+b{)}^n=su{m}_{k=0}^{n}binomnk{a}^{n-k}{b}^k(a\; +\; b)^n\; =\; \backslash \backslash sum\_\{k=0\}^\{n\}\; \backslash \backslash binom\{n\}\{k\}\; a^\{n\; -\; k\}\; b^k$. Here, $a=frac35xa\; =\; \backslash \backslash frac\{3\}\{5\}x$, $b=-fracy2b\; =\; -\backslash \backslash frac\{y\}\{2\}$, and $n=8n\; =\; 8$.\n\nWe need the term where the exponents of $xx$ and $yy$ are 2 and 6, respectively. This corresponds to $k=6k\; =\; 6$ because the exponent on $xx$ is $8-k=28\; -\; k\; =\; 2$ and the exponent on $yy$ is $k=6k\; =\; 6$.\n\nThe specific term is:\n\\[\n\\binom{8}{6} \\left(\\frac{3}{5}x\\right)^2 \\left(-\\frac{y}{2}\\right)^6\n\\]\n\nFirst, compute the binomial coefficient:\n\\[\n\\binom{8}{6} = \\frac{8!}{6! \\cdot 2!} = \\frac{40320}{720 \\cdot 2} = 28\n\\]\n\nNext, compute $left(frac35right{)}^2\backslash \backslash left(\backslash \backslash frac\{3\}\{5\}\backslash \backslash right)^2$:\n\\[\n\\left(\\frac{3}{5}\\right)^2 = \\frac{9}{25}\n\\]\n\nThen, compute $left(-frac12right{)}^6\backslash \backslash left(-\backslash \backslash frac\{1\}\{2\}\backslash \backslash right)^6$:\n\\[\n\\left(-\\frac{1}{2}\\right)^6 = \\frac{1}{64}\n\\]\n\nPutting it all together, the term is:\n\\[\n28 \\cdot \\frac{9}{25} \\cdot \\frac{1}{64}\n\\]\n\nMultiply the numerical coefficients:\n\\[\n28 \\times \\frac{9}{25} \\times \\frac{1}{64} = \\frac{28 \\times 9 \\times 1}{25 \\times 64} = \\frac{252}{1600}\n\\]\n\nSimplify the fraction:\n\\[\n\\frac{252}{1600} = \\frac{63}{400}\n\\]\n\nThus, the coefficient of $x^2{y}^{6}x^2\; y^6$ is $boxeddfrac63400\backslash \backslash boxed\{\backslash \backslash dfrac\{63\}\{400\}\}$.",
    "messages": [
        {
            "content": "What is the coefficient of $x^2y^6$ in the expansion of $\\left(\\frac{3}{5}x-\\frac{y}{2}\\right)^8$?  Express your answer as a common fraction.",
            "role": "user"
        },
        {
            "content": "To determine the coefficient of $x^2{y}^{6}x^2y^6$ in the expansion of $left(frac35x-fracy2right{)}^8\backslash \backslash left(\backslash \backslash frac\{3\}\{5\}x\; -\; \backslash \backslash frac\{y\}\{2\}\backslash \backslash right)^8$, we can use the binomial theorem.\n\nThe binomial theorem states:\n\\[\n(a + b)^n = \\sum_{k=0}^{n} \\binom{n}{k} a^{n-k} b^k\n\\]\n\nIn this case, $a=frac35xa\; =\; \backslash \backslash frac\{3\}\{5\}x$, $b=-fracy2b\; =\; -\backslash \backslash frac\{y\}\{2\}$, and $n=8n\; =\; 8$.\n\nWe are interested in the term that contains $x^2{y}^{6}x^2y^6$. In the general term of the binomial expansion:\n\\[\n\\binom{8}{k} \\left(\\frac{3}{5}x\\right)^{8-k} \\left(-\\frac{y}{2}\\right)^k\n\\]\n\nTo get $x^{2}x^2$, we need $8-k=28\; -\; k\; =\; 2$, thus $k=6k\; =\; 6$.\n\nSubstituting $k=6k\; =\; 6$ into the expression:\n\\[\n\\binom{8}{6} \\left(\\frac{3}{5}x\\right)^{8-6} \\left(-\\frac{y}{2}\\right)^6 = \\binom{8}{6} \\left(\\frac{3}{5}x\\right)^2 \\left(-\\frac{y}{2}\\right)^6\n\\]\n\nNow, we will compute each part of this expression.\n\n1. Calculate the binomial coefficient $binom86\backslash \backslash binom\{8\}\{6\}$.\n2. Compute $left(frac35right{)}^2\backslash \backslash left(\backslash \backslash frac\{3\}\{5\}\backslash \backslash right)^2$.\n3. Compute $left(-fracy2right{)}^6\backslash \backslash left(-\backslash \backslash frac\{y\}\{2\}\backslash \backslash right)^6$.\n4. Combine everything together to get the coefficient of $x^2{y}^{6}x^2y^6$.\n\nLet\u0027s compute these in Python.\n```python\nfrom math import comb\n\n# Given values\nn = 8\nk = 6\n\n# Calculate the binomial coefficient\nbinom_coeff = comb(n, k)\n\n# Compute (3/5)^2\na_term = (3/5)**2\n\n# Compute (-1/2)^6\nb_term = (-1/2)**6\n\n# Combine terms to get the coefficient of x^2y^6\ncoefficient = binom_coeff * a_term * b_term\nprint(coefficient)\n```\n```output\n0.1575\n```\nThe coefficient of $x^2{y}^{6}x^2y^6$ in the expansion of $left(frac35x-fracy2right{)}^8\backslash \backslash left(\backslash \backslash frac\{3\}\{5\}x\; -\; \backslash \backslash frac\{y\}\{2\}\backslash \backslash right)^8$ is $0.15750.1575$. To express this as a common fraction, we recognize that:\n\n\\[ 0.1575 = \\frac{1575}{10000} = \\frac{63}{400} \\]\n\nThus, the coefficient can be expressed as:\n\n\\[\n\\boxed{\\frac{63}{400}}\n\\]",
            "role": "assistant"
        }
    ],
    "model_name": "/data/llm_models/deepseek-ai/DeepSeek-R1-Distill-Qwen-14B",
    "problem": "What is the coefficient of $x^2y^6$ in the expansion of $\\left(\\frac{3}{5}x-\\frac{y}{2}\\right)^8$?  Express your answer as a common fraction.",
    "solution": "To determine the coefficient of $x^2{y}^{6}x^2y^6$ in the expansion of $left(frac35x-fracy2right{)}^8\backslash \backslash left(\backslash \backslash frac\{3\}\{5\}x\; -\; \backslash \backslash frac\{y\}\{2\}\backslash \backslash right)^8$, we can use the binomial theorem.\n\nThe binomial theorem states:\n\\[\n(a + b)^n = \\sum_{k=0}^{n} \\binom{n}{k} a^{n-k} b^k\n\\]\n\nIn this case, $a=frac35xa\; =\; \backslash \backslash frac\{3\}\{5\}x$, $b=-fracy2b\; =\; -\backslash \backslash frac\{y\}\{2\}$, and $n=8n\; =\; 8$.\n\nWe are interested in the term that contains $x^2{y}^{6}x^2y^6$. In the general term of the binomial expansion:\n\\[\n\\binom{8}{k} \\left(\\frac{3}{5}x\\right)^{8-k} \\left(-\\frac{y}{2}\\right)^k\n\\]\n\nTo get $x^{2}x^2$, we need $8-k=28\; -\; k\; =\; 2$, thus $k=6k\; =\; 6$.\n\nSubstituting $k=6k\; =\; 6$ into the expression:\n\\[\n\\binom{8}{6} \\left(\\frac{3}{5}x\\right)^{8-6} \\left(-\\frac{y}{2}\\right)^6 = \\binom{8}{6} \\left(\\frac{3}{5}x\\right)^2 \\left(-\\frac{y}{2}\\right)^6\n\\]\n\nNow, we will compute each part of this expression.\n\n1. Calculate the binomial coefficient $binom86\backslash \backslash binom\{8\}\{6\}$.\n2. Compute $left(frac35right{)}^2\backslash \backslash left(\backslash \backslash frac\{3\}\{5\}\backslash \backslash right)^2$.\n3. Compute $left(-fracy2right{)}^6\backslash \backslash left(-\backslash \backslash frac\{y\}\{2\}\backslash \backslash right)^6$.\n4. Combine everything together to get the coefficient of $x^2{y}^{6}x^2y^6$.\n\nLet\u0027s compute these in Python.\n```python\nfrom math import comb\n\n# Given values\nn = 8\nk = 6\n\n# Calculate the binomial coefficient\nbinom_coeff = comb(n, k)\n\n# Compute (3/5)^2\na_term = (3/5)**2\n\n# Compute (-1/2)^6\nb_term = (-1/2)**6\n\n# Combine terms to get the coefficient of x^2y^6\ncoefficient = binom_coeff * a_term * b_term\nprint(coefficient)\n```\n```output\n0.1575\n```\nThe coefficient of $x^2{y}^{6}x^2y^6$ in the expansion of $left(frac35x-fracy2right{)}^8\backslash \backslash left(\backslash \backslash frac\{3\}\{5\}x\; -\; \backslash \backslash frac\{y\}\{2\}\backslash \backslash right)^8$ is $0.15750.1575$. To express this as a common fraction, we recognize that:\n\n\\[ 0.1575 = \\frac{1575}{10000} = \\frac{63}{400} \\]\n\nThus, the coefficient can be expressed as:\n\n\\[\n\\boxed{\\frac{63}{400}}\n\\]"
}

This subset can be loaded as:

from datasets import load_dataset

ds = load_dataset("mikehps123/numina-deepseek-r1-qwen-14b", "default")

Or simply as it follows, since there's only one configuration and is named default:

from datasets import load_dataset

ds = load_dataset("mikehps123/numina-deepseek-r1-qwen-14b")