size_categories:n<1K format:parquet modality:text library:datasets library:pandas library:mlcroissant library:polars library:distilabel region:us synthetic distilabel rlaif
---
size_categories: n<1K
dataset_info:
  features:
  - name: problem
    dtype: string
  - name: solution
    dtype: string
  - name: messages
    list:
    - name: content
      dtype: string
    - name: role
      dtype: string
  - name: generation
    dtype: string
  - name: distilabel_metadata
    struct:
    - name: raw_input_text_generation_0
      list:
      - name: content
        dtype: string
      - name: role
        dtype: string
    - name: raw_output_text_generation_0
      dtype: string
    - name: statistics_text_generation_0
      struct:
      - name: input_tokens
        dtype: int64
      - name: output_tokens
        dtype: int64
  - name: model_name
    dtype: string
  splits:
  - name: train
    num_bytes: 841992
    num_examples: 40
  download_size: 277296
  dataset_size: 841992
configs:
- config_name: default
  data_files:
  - split: train
    path: data/train-*
tags:
- synthetic
- distilabel
- rlaif
---

<p align="left">
  <a href="https://github.com/argilla-io/distilabel">
    <img src="https://raw.githubusercontent.com/argilla-io/distilabel/main/docs/assets/distilabel-badge-light.png" alt="Built with Distilabel" width="200" height="32"/>
  </a>
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# Dataset Card for numina-deepseek-r1-qwen-7b

This dataset has been created with [distilabel](https://distilabel.argilla.io/).



## 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:

```console
distilabel pipeline run --config "https://huggingface.co/datasets/alfredcs/numina-deepseek-r1-qwen-7b/raw/main/pipeline.yaml"
```

or explore the configuration:

```console
distilabel pipeline info --config "https://huggingface.co/datasets/alfredcs/numina-deepseek-r1-qwen-7b/raw/main/pipeline.yaml"
```

## Dataset structure

The examples have the following structure per configuration:


<details><summary> Configuration: default </summary><hr>

```json
{
    "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": "\u003cthink\u003e\nAlright, so I need to find the coefficient of \\(x^2y^6\\) in the expansion of \\(\\left(\\frac{3}{5}x - \\frac{y}{2}\\right)^8\\). Hmm, binomial expansion, right? Let me think about how to approach this.\n\nFirst, I remember that the binomial theorem allows us to expand expressions of the form \\((a + b)^n\\). The general term in the expansion is given by \\(\\binom{n}{k}a^{n - k}b^k\\), where \\(k\\) ranges from 0 to \\(n\\). In this case, my expression is \\(\\left(\\frac{3}{5}x - \\frac{y}{2}\\right)^8\\), so \\(a = \\frac{3}{5}x\\) and \\(b = -\\frac{y}{2}\\), and \\(n = 8\\).\n\nI need the term where the power of \\(x\\) is 2 and the power of \\(y\\) is 6. That means I need to find the term where \\(x^2y^6\\) appears. Since the exponents in each term must add up to 8 (because of the exponent on the binomial), if \\(x\\) is squared, then \\(y\\) must be to the 6th power, which adds up to 8. Perfect, that fits.\n\nSo, to find this specific term, I can use the binomial formula. The term will be \\(\\binom{8}{k} \\left(\\frac{3}{5}x\\right)^{8 - k} \\left(-\\frac{y}{2}\\right)^k\\). I need to find the value of \\(k\\) such that the exponents on \\(x\\) and \\(y\\) are 2 and 6, respectively.\n\nLooking at the exponents, since \\(x\\) is raised to the power of \\(8 - k\\), I want that to be 2. So, \\(8 - k = 2\\), which means \\(k = 6\\). Perfect, so \\(k = 6\\).\n\nNow, substituting \\(k = 6\\) into the term, we get:\n\n\\[\n\\binom{8}{6} \\left(\\frac{3}{5}x\\right)^{2} \\left(-\\frac{y}{2}\\right)^6\n\\]\n\nLet me compute each part step by step.\n\nFirst, the binomial coefficient \\(\\binom{8}{6}\\). Remember that \\(\\binom{n}{k}\\) is equal to \\(\\frac{n!}{k!(n - k)!}\\). So,\n\n\\[\n\\binom{8}{6} = \\frac{8!}{6!2!} = \\frac{8 \\times 7 \\times 6!}{6! \\times 2 \\times 1} = \\frac{8 \\times 7}{2 \\times 1} = \\frac{56}{2} = 28\n\\]\n\nOkay, so that\u0027s 28.\n\nNext, let\u0027s compute \\(\\left(\\frac{3}{5}x\\right)^2\\). Squaring both the coefficient and the variable part:\n\n\\[\n\\left(\\frac{3}{5}\\right)^2 x^2 = \\frac{9}{25}x^2\n\\]\n\nGood, so that\u0027s \\(\\frac{9}{25}x^2\\).\n\nNow, the other part is \\(\\left(-\\frac{y}{2}\\right)^6\\). Let\u0027s compute that.\n\nFirst, note that when you raise a negative number to an even power, the result is positive. So, \\(-1\\) raised to the 6th power is 1. Then,\n\n\\[\n\\left(-\\frac{1}{2}\\right)^6 = \\left(\\frac{1}{2}\\right)^6 = \\frac{1}{64}\n\\]\n\nBut wait, let me do it step by step:\n\n\\[\n\\left(-\\frac{y}{2}\\right)^6 = (-1)^6 \\times \\left(\\frac{y}{2}\\right)^6 = 1 \\times \\frac{y^6}{64} = \\frac{y^6}{64}\n\\]\n\nSo, that\u0027s \\(\\frac{y^6}{64}\\).\n\nPutting it all together, the term is:\n\n\\[\n28 \\times \\frac{9}{25}x^2 \\times \\frac{1}{64}y^6\n\\]\n\nNow, let\u0027s multiply the constants together:\n\nFirst, multiply 28 and \\(\\frac{9}{25}\\):\n\n\\[\n28 \\times \\frac{9}{25} = \\frac{28 \\times 9}{25} = \\frac{252}{25}\n\\]\n\nThen, multiply that result by \\(\\frac{1}{64}\\):\n\n\\[\n\\frac{252}{25} \\times \\frac{1}{64} = \\frac{252}{1600}\n\\]\n\nNow, let\u0027s simplify \\(\\frac{252}{1600}\\). To do this, I can find the greatest common divisor (GCD) of 252 and 1600.\n\nFirst, prime factorization of 252:\n\n252 \u00f7 2 = 126\n\n126 \u00f7 2 = 63\n\n63 \u00f7 3 = 21\n\n21 \u00f7 3 = 7\n\n7 is prime.\n\nSo, 252 = 2^2 \u00d7 3^2 \u00d7 7\n\nNow, prime factorization of 1600:\n\n1600 \u00f7 2 = 800\n\n800 \u00f7 2 = 400\n\n400 \u00f7 2 = 200\n\n200 \u00f7 2 = 100\n\n100 \u00f7 2 = 50\n\n50 \u00f7 2 = 25\n\n25 \u00f7 5 = 5\n\n5 \u00f7 5 = 1\n\nSo, 1600 = 2^6 \u00d7 5^2\n\nNow, the GCD is the minimum exponents of the common prime factors.\n\nCommon primes between 252 and 1600 are only 2.\n\nMinimum exponent of 2 is 2 (since 252 has 2^2 and 1600 has 2^6). So, GCD is 2^2 = 4.\n\nSo, let\u0027s divide numerator and denominator by 4:\n\n252 \u00f7 4 = 63\n\n1600 \u00f7 4 = 400\n\nSo, \\(\\frac{252}{1600}\\) simplifies to \\(\\frac{63}{400}\\).\n\nTherefore, the coefficient is \\(\\frac{63}{400}\\).\n\nWait, let me double-check to see if I can reduce it further. 63 and 400 have no common factors besides 1, since 63 is 7\u00d79 and 400 is 16\u00d725. So, yes, \\(\\frac{63}{400}\\) is in simplest form.\n\nSo, putting it all together, the term is:\n\n\\[\n\\frac{63}{400}x^2y^6\n\\]\n\nTherefore, the coefficient of \\(x^2y^6\\) is \\(\\frac{63}{400}\\).\n\nWait, let me just go through the steps again to make sure I didn\u0027t make a mistake.\n\n1. Binomial coefficient: \\(\\binom{8}{6} = 28\\). That seems right because \\(\\binom{8}{6} = \\binom{8}{2}\\) since \\(\\binom{n}{k} = \\binom{n}{n - k}\\), and \\(\\binom{8}{2} = \\frac{8\u00d77}{2} = 28\\). Correct.\n\n2. \\(\\left(\\frac{3}{5}\\right)^2 = \\frac{9}{25}\\). Correct.\n\n3. \\(\\left(-\\frac{1}{2}\\right)^6 = \\frac{1}{64}\\). Since the exponent is even, it\u0027s positive. Correct.\n\n4. Multiply constants: 28 \u00d7 9/25 \u00d7 1/64.\n\nFirst, 28 \u00d7 9 = 252, correct.\n\nThen, 252 \u00f7 25 = 10.08, but we keep it as fractions.\n\n252/25 \u00d7 1/64 = 252/(25\u00d764) = 252/1600.\n\nSimplify 252/1600: divide numerator and denominator by 4: 63/400. Correct.\n\nSo, yes, the coefficient is 63/400.\n\nI think that\u0027s correct. I don\u0027t see any mistakes in the calculations.\n\n**Final Answer**\nThe coefficient of \\(x^2y^6\\) is \\boxed{\\dfrac{63}{400}}.\n\u003c/think\u003e\n\nTo find the coefficient of \\(x^2y^6\\) in the expansion of \\(\\left(\\frac{3}{5}x - \\frac{y}{2}\\right)^8\\), we use the binomial theorem. The general term in the expansion is given by:\n\n\\[\n\\binom{n}{k}a^{n - k}b^k\n\\]\n\nwhere \\(a = \\frac{3}{5}x\\), \\(b = -\\frac{y}{2}\\), and \\(n = 8\\). We need the term where the exponents of \\(x\\) and \\(y\\) are 2 and 6, respectively. This corresponds to \\(k = 6\\) because \\(8 - k = 2\\).\n\nThe term is:\n\n\\[\n\\binom{8}{6} \\left(\\frac{3}{5}x\\right)^2 \\left(-\\frac{y}{2}\\right)^6\n\\]\n\nFirst, we compute the binomial coefficient:\n\n\\[\n\\binom{8}{6} = \\frac{8!}{6!2!} = \\frac{8 \\times 7}{2 \\times 1} = 28\n\\]\n\nNext, we compute the powers:\n\n\\[\n\\left(\\frac{3}{5}\\right)^2 = \\frac{9}{25}\n\\]\n\n\\[\n\\left(-\\frac{1}{2}\\right)^6 = \\frac{1}{64}\n\\]\n\nMultiplying these together with the binomial coefficient:\n\n\\[\n28 \\times \\frac{9}{25} \\times \\frac{1}{64}\n\\]\n\nFirst, multiply 28 and \\(\\frac{9}{25}\\):\n\n\\[\n28 \\times \\frac{9}{25} = \\frac{252}{25}\n\\]\n\nThen, multiply by \\(\\frac{1}{64}\\):\n\n\\[\n\\frac{252}{25} \\times \\frac{1}{64} = \\frac{252}{1600}\n\\]\n\nSimplifying \\(\\frac{252}{1600}\\) by dividing both numerator and denominator by their GCD (which is 4):\n\n\\[\n\\frac{252}{1600} = \\frac{63}{400}\n\\]\n\nThus, the coefficient of \\(x^2y^6\\) is \\(\\boxed{\\dfrac{63}{400}}\\).",
        "statistics_text_generation_0": {
            "input_tokens": 77,
            "output_tokens": 2543
        }
    },
    "generation": "\u003cthink\u003e\nAlright, so I need to find the coefficient of \\(x^2y^6\\) in the expansion of \\(\\left(\\frac{3}{5}x - \\frac{y}{2}\\right)^8\\). Hmm, binomial expansion, right? Let me think about how to approach this.\n\nFirst, I remember that the binomial theorem allows us to expand expressions of the form \\((a + b)^n\\). The general term in the expansion is given by \\(\\binom{n}{k}a^{n - k}b^k\\), where \\(k\\) ranges from 0 to \\(n\\). In this case, my expression is \\(\\left(\\frac{3}{5}x - \\frac{y}{2}\\right)^8\\), so \\(a = \\frac{3}{5}x\\) and \\(b = -\\frac{y}{2}\\), and \\(n = 8\\).\n\nI need the term where the power of \\(x\\) is 2 and the power of \\(y\\) is 6. That means I need to find the term where \\(x^2y^6\\) appears. Since the exponents in each term must add up to 8 (because of the exponent on the binomial), if \\(x\\) is squared, then \\(y\\) must be to the 6th power, which adds up to 8. Perfect, that fits.\n\nSo, to find this specific term, I can use the binomial formula. The term will be \\(\\binom{8}{k} \\left(\\frac{3}{5}x\\right)^{8 - k} \\left(-\\frac{y}{2}\\right)^k\\). I need to find the value of \\(k\\) such that the exponents on \\(x\\) and \\(y\\) are 2 and 6, respectively.\n\nLooking at the exponents, since \\(x\\) is raised to the power of \\(8 - k\\), I want that to be 2. So, \\(8 - k = 2\\), which means \\(k = 6\\). Perfect, so \\(k = 6\\).\n\nNow, substituting \\(k = 6\\) into the term, we get:\n\n\\[\n\\binom{8}{6} \\left(\\frac{3}{5}x\\right)^{2} \\left(-\\frac{y}{2}\\right)^6\n\\]\n\nLet me compute each part step by step.\n\nFirst, the binomial coefficient \\(\\binom{8}{6}\\). Remember that \\(\\binom{n}{k}\\) is equal to \\(\\frac{n!}{k!(n - k)!}\\). So,\n\n\\[\n\\binom{8}{6} = \\frac{8!}{6!2!} = \\frac{8 \\times 7 \\times 6!}{6! \\times 2 \\times 1} = \\frac{8 \\times 7}{2 \\times 1} = \\frac{56}{2} = 28\n\\]\n\nOkay, so that\u0027s 28.\n\nNext, let\u0027s compute \\(\\left(\\frac{3}{5}x\\right)^2\\). Squaring both the coefficient and the variable part:\n\n\\[\n\\left(\\frac{3}{5}\\right)^2 x^2 = \\frac{9}{25}x^2\n\\]\n\nGood, so that\u0027s \\(\\frac{9}{25}x^2\\).\n\nNow, the other part is \\(\\left(-\\frac{y}{2}\\right)^6\\). Let\u0027s compute that.\n\nFirst, note that when you raise a negative number to an even power, the result is positive. So, \\(-1\\) raised to the 6th power is 1. Then,\n\n\\[\n\\left(-\\frac{1}{2}\\right)^6 = \\left(\\frac{1}{2}\\right)^6 = \\frac{1}{64}\n\\]\n\nBut wait, let me do it step by step:\n\n\\[\n\\left(-\\frac{y}{2}\\right)^6 = (-1)^6 \\times \\left(\\frac{y}{2}\\right)^6 = 1 \\times \\frac{y^6}{64} = \\frac{y^6}{64}\n\\]\n\nSo, that\u0027s \\(\\frac{y^6}{64}\\).\n\nPutting it all together, the term is:\n\n\\[\n28 \\times \\frac{9}{25}x^2 \\times \\frac{1}{64}y^6\n\\]\n\nNow, let\u0027s multiply the constants together:\n\nFirst, multiply 28 and \\(\\frac{9}{25}\\):\n\n\\[\n28 \\times \\frac{9}{25} = \\frac{28 \\times 9}{25} = \\frac{252}{25}\n\\]\n\nThen, multiply that result by \\(\\frac{1}{64}\\):\n\n\\[\n\\frac{252}{25} \\times \\frac{1}{64} = \\frac{252}{1600}\n\\]\n\nNow, let\u0027s simplify \\(\\frac{252}{1600}\\). To do this, I can find the greatest common divisor (GCD) of 252 and 1600.\n\nFirst, prime factorization of 252:\n\n252 \u00f7 2 = 126\n\n126 \u00f7 2 = 63\n\n63 \u00f7 3 = 21\n\n21 \u00f7 3 = 7\n\n7 is prime.\n\nSo, 252 = 2^2 \u00d7 3^2 \u00d7 7\n\nNow, prime factorization of 1600:\n\n1600 \u00f7 2 = 800\n\n800 \u00f7 2 = 400\n\n400 \u00f7 2 = 200\n\n200 \u00f7 2 = 100\n\n100 \u00f7 2 = 50\n\n50 \u00f7 2 = 25\n\n25 \u00f7 5 = 5\n\n5 \u00f7 5 = 1\n\nSo, 1600 = 2^6 \u00d7 5^2\n\nNow, the GCD is the minimum exponents of the common prime factors.\n\nCommon primes between 252 and 1600 are only 2.\n\nMinimum exponent of 2 is 2 (since 252 has 2^2 and 1600 has 2^6). So, GCD is 2^2 = 4.\n\nSo, let\u0027s divide numerator and denominator by 4:\n\n252 \u00f7 4 = 63\n\n1600 \u00f7 4 = 400\n\nSo, \\(\\frac{252}{1600}\\) simplifies to \\(\\frac{63}{400}\\).\n\nTherefore, the coefficient is \\(\\frac{63}{400}\\).\n\nWait, let me double-check to see if I can reduce it further. 63 and 400 have no common factors besides 1, since 63 is 7\u00d79 and 400 is 16\u00d725. So, yes, \\(\\frac{63}{400}\\) is in simplest form.\n\nSo, putting it all together, the term is:\n\n\\[\n\\frac{63}{400}x^2y^6\n\\]\n\nTherefore, the coefficient of \\(x^2y^6\\) is \\(\\frac{63}{400}\\).\n\nWait, let me just go through the steps again to make sure I didn\u0027t make a mistake.\n\n1. Binomial coefficient: \\(\\binom{8}{6} = 28\\). That seems right because \\(\\binom{8}{6} = \\binom{8}{2}\\) since \\(\\binom{n}{k} = \\binom{n}{n - k}\\), and \\(\\binom{8}{2} = \\frac{8\u00d77}{2} = 28\\). Correct.\n\n2. \\(\\left(\\frac{3}{5}\\right)^2 = \\frac{9}{25}\\). Correct.\n\n3. \\(\\left(-\\frac{1}{2}\\right)^6 = \\frac{1}{64}\\). Since the exponent is even, it\u0027s positive. Correct.\n\n4. Multiply constants: 28 \u00d7 9/25 \u00d7 1/64.\n\nFirst, 28 \u00d7 9 = 252, correct.\n\nThen, 252 \u00f7 25 = 10.08, but we keep it as fractions.\n\n252/25 \u00d7 1/64 = 252/(25\u00d764) = 252/1600.\n\nSimplify 252/1600: divide numerator and denominator by 4: 63/400. Correct.\n\nSo, yes, the coefficient is 63/400.\n\nI think that\u0027s correct. I don\u0027t see any mistakes in the calculations.\n\n**Final Answer**\nThe coefficient of \\(x^2y^6\\) is \\boxed{\\dfrac{63}{400}}.\n\u003c/think\u003e\n\nTo find the coefficient of \\(x^2y^6\\) in the expansion of \\(\\left(\\frac{3}{5}x - \\frac{y}{2}\\right)^8\\), we use the binomial theorem. The general term in the expansion is given by:\n\n\\[\n\\binom{n}{k}a^{n - k}b^k\n\\]\n\nwhere \\(a = \\frac{3}{5}x\\), \\(b = -\\frac{y}{2}\\), and \\(n = 8\\). We need the term where the exponents of \\(x\\) and \\(y\\) are 2 and 6, respectively. This corresponds to \\(k = 6\\) because \\(8 - k = 2\\).\n\nThe term is:\n\n\\[\n\\binom{8}{6} \\left(\\frac{3}{5}x\\right)^2 \\left(-\\frac{y}{2}\\right)^6\n\\]\n\nFirst, we compute the binomial coefficient:\n\n\\[\n\\binom{8}{6} = \\frac{8!}{6!2!} = \\frac{8 \\times 7}{2 \\times 1} = 28\n\\]\n\nNext, we compute the powers:\n\n\\[\n\\left(\\frac{3}{5}\\right)^2 = \\frac{9}{25}\n\\]\n\n\\[\n\\left(-\\frac{1}{2}\\right)^6 = \\frac{1}{64}\n\\]\n\nMultiplying these together with the binomial coefficient:\n\n\\[\n28 \\times \\frac{9}{25} \\times \\frac{1}{64}\n\\]\n\nFirst, multiply 28 and \\(\\frac{9}{25}\\):\n\n\\[\n28 \\times \\frac{9}{25} = \\frac{252}{25}\n\\]\n\nThen, multiply by \\(\\frac{1}{64}\\):\n\n\\[\n\\frac{252}{25} \\times \\frac{1}{64} = \\frac{252}{1600}\n\\]\n\nSimplifying \\(\\frac{252}{1600}\\) by dividing both numerator and denominator by their GCD (which is 4):\n\n\\[\n\\frac{252}{1600} = \\frac{63}{400}\n\\]\n\nThus, the coefficient of \\(x^2y^6\\) is \\(\\boxed{\\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^2y^6\\) in the expansion of \\(\\left(\\frac{3}{5}x - \\frac{y}{2}\\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 = \\frac{3}{5}x\\), \\(b = -\\frac{y}{2}\\), and \\(n = 8\\).\n\nWe are interested in the term that contains \\(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\\), we need \\(8 - k = 2\\), thus \\(k = 6\\).\n\nSubstituting \\(k = 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 \\(\\binom{8}{6}\\).\n2. Compute \\(\\left(\\frac{3}{5}\\right)^2\\).\n3. Compute \\(\\left(-\\frac{y}{2}\\right)^6\\).\n4. Combine everything together to get the coefficient of \\(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^2y^6\\) in the expansion of \\(\\left(\\frac{3}{5}x - \\frac{y}{2}\\right)^8\\) is \\(0.1575\\). 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": "deepseek-ai/DeepSeek-R1-Distill-Qwen-7B",
    "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^2y^6\\) in the expansion of \\(\\left(\\frac{3}{5}x - \\frac{y}{2}\\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 = \\frac{3}{5}x\\), \\(b = -\\frac{y}{2}\\), and \\(n = 8\\).\n\nWe are interested in the term that contains \\(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\\), we need \\(8 - k = 2\\), thus \\(k = 6\\).\n\nSubstituting \\(k = 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 \\(\\binom{8}{6}\\).\n2. Compute \\(\\left(\\frac{3}{5}\\right)^2\\).\n3. Compute \\(\\left(-\\frac{y}{2}\\right)^6\\).\n4. Combine everything together to get the coefficient of \\(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^2y^6\\) in the expansion of \\(\\left(\\frac{3}{5}x - \\frac{y}{2}\\right)^8\\) is \\(0.1575\\). 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:

```python
from datasets import load_dataset

ds = load_dataset("alfredcs/numina-deepseek-r1-qwen-7b", "default")
```

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

```python
from datasets import load_dataset

ds = load_dataset("alfredcs/numina-deepseek-r1-qwen-7b")
```


</details>




