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: 793700
    num_examples: 40
  download_size: 276335
  dataset_size: 793700
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>
</p>

# 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/3482brother/numina-deepseek-r1-qwen-7b/raw/main/pipeline.yaml"
```

or explore the configuration:

```console
distilabel pipeline info --config "https://huggingface.co/datasets/3482brother/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 in brackets: [Final Answer]\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 have this problem: I need to find the coefficient of \\( x^2 y^6 \\) in the expansion of \\( \\left( \\frac{3}{5}x - \\frac{y}{2} \\right)^8 \\). Hmm, okay. I remember that when expanding binomials, the general term can be found using the binomial theorem. Let me recall what that is.\n\nThe binomial theorem states that \\( (a + b)^n = \\sum_{k=0}^{n} \\binom{n}{k} a^{n - k} b^k \\). So, in this case, my \\( a \\) is \\( \\frac{3}{5}x \\) and my \\( b \\) is \\( -\\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. Let me think about how that relates to the exponents in the binomial theorem. Since the term \\( x^2 y^6 \\) implies that \\( x \\) is raised to the 2nd power and \\( y \\) is raised to the 6th power, that should correspond to \\( a^{2} b^{6} \\) in the expansion.\n\nWait, let me check that: since \\( a = \\frac{3}{5}x \\), \\( a^2 \\) would be \\( \\left( \\frac{3}{5}x \\right)^2 = \\left( \\frac{3}{5} \\right)^2 x^2 \\). Similarly, \\( b = -\\frac{y}{2} \\), so \\( b^6 = \\left( -\\frac{y}{2} \\right)^6 \\). Hmm, okay, that makes sense.\n\nSo, the general term is \\( \\binom{8}{k} a^{8 - k} b^{k} \\). I need the term where the power of \\( x \\) is 2 and the power of \\( y \\) is 6. Since \\( a \\) contributes the \\( x \\) term and \\( b \\) contributes the \\( y \\) term, I can set up the exponents accordingly.\n\nLet me denote the exponent of \\( a \\) as \\( 8 - k \\) and the exponent of \\( b \\) as \\( k \\). Since \\( a \\) is \\( x \\), its exponent \\( 8 - k \\) should be equal to 2, right? So, \\( 8 - k = 2 \\) which implies \\( k = 6 \\). So, the term we\u0027re interested in is when \\( k = 6 \\).\n\nTherefore, the term is \\( \\binom{8}{6} a^{2} b^{6} \\). Let me compute each part step by step.\n\nFirst, compute the binomial coefficient \\( \\binom{8}{6} \\). I remember that \\( \\binom{n}{k} = \\frac{n!}{k!(n - k)!} \\). So, plugging in the values:\n\n\\( \\binom{8}{6} = \\frac{8!}{6! \\cdot (8 - 6)!} = \\frac{8!}{6! \\cdot 2!} \\). Calculating that:\n\n\\( 8! = 40320 \\), \\( 6! = 720 \\), \\( 2! = 2 \\).\n\nSo, \\( \\frac{40320}{720 \\cdot 2} = \\frac{40320}{1440} \\). Let me compute that: 40320 divided by 1440. Hmm, 1440 times 28 is 40320 because 1440 times 10 is 14400, 1440 times 20 is 28800, and 1440 times 8 is 11520; adding 28800 + 11520 gives 40320. So, 1440 times 28 is 40320, so \\( \\frac{40320}{1440} = 28 \\). So, the binomial coefficient is 28.\n\nNext, compute \\( a^2 \\): \\( a = \\frac{3}{5}x \\), so \\( a^2 = \\left( \\frac{3}{5} \\right)^2 x^2 = \\frac{9}{25}x^2 \\).\n\nThen, compute \\( b^6 \\): \\( b = -\\frac{y}{2} \\), so \\( b^6 = \\left( -\\frac{1}{2} \\right)^6 y^6 \\). Wait, hold on, \\( \\left( -\\frac{1}{2} \\right)^6 \\) is the same as \\( \\left( \\frac{1}{2} \\right)^6 \\) because any even power of a negative number is positive. So, \\( \\left( \\frac{1}{2} \\right)^6 = \\frac{1}{64} \\). Therefore, \\( b^6 = \\frac{1}{64} y^6 \\).\n\nNow, multiply all these parts together: the binomial coefficient, \\( a^2 \\), and \\( b^6 \\).\n\nSo, putting it all together:\n\nTerm = \\( \\binom{8}{6} \\cdot a^2 \\cdot b^6 = 28 \\cdot \\frac{9}{25}x^2 \\cdot \\frac{1}{64}y^6 \\).\n\nNow, let\u0027s multiply the constants:\n\nFirst, multiply 28 and \\( \\frac{9}{25} \\):\n\n28 times \\( \\frac{9}{25} \\) is \\( \\frac{28 \\times 9}{25} \\).\n\nCompute 28 times 9: 28 times 10 is 280, minus 28 is 252. So, \\( \\frac{252}{25} \\).\n\nThen, multiply that by \\( \\frac{1}{64} \\):\n\nSo, \\( \\frac{252}{25} \\times \\frac{1}{64} = \\frac{252}{1600} \\).\n\nSimplify \\( \\frac{252}{1600} \\). Let\u0027s see.\n\nDivide numerator and denominator by 4: 252 \u00f7 4 = 63, 1600 \u00f7 4 = 400.\n\nSo, \\( \\frac{63}{400} \\).\n\nIs that reducible? Let\u0027s check. 63 factors are 7 and 9, 400 factors are 2^4 and 5^2. No common factors, so \\( \\frac{63}{400} \\) is the simplified form.\n\nSo, the coefficient is \\( \\frac{63}{400} \\). Therefore, the coefficient of \\( x^2 y^6 \\) is \\( \\frac{63}{400} \\).\n\nWait a second, let me double-check my calculations to make sure I didn\u0027t make a mistake.\n\nSo, binomial coefficient: 8 choose 6 is 28, that\u0027s correct.\n\n\\( a^2 = \\left( \\frac{3}{5} \\right)^2 = \\frac{9}{25} \\), that\u0027s right.\n\n\\( b^6 = \\left( -\\frac{1}{2} \\right)^6 = \\frac{1}{64} \\), correct.\n\nMultiply 28 * 9 = 252, yes, and 252 divided by 25 * 64. Wait, no: 28 * 9 / (25 * 64). Wait, that\u0027s 28 * 9 = 252, and 25 * 64 = 1600, so 252 / 1600. Then, simplifying, divide numerator and denominator by 4: 63 / 400, which is 0.1575.\n\nIs there a simpler way to compute the coefficient? Let me see.\n\nAlternatively, 28 * 9 = 252, 252 / (25 * 64). Since 25 * 64 is 1600, so 252 / 1600. 252 divided by 4 is 63, 1600 divided by 4 is 400. So, yeah, 63/400. That\u0027s correct.\n\nWait, but let me think again about the signs. \\( b = -\\frac{y}{2} \\), so \\( b^6 = \\left( -\\frac{1}{2} \\right)^6 = \\frac{1}{64} \\), because the exponent is even. So, the sign is positive. Therefore, the coefficient is indeed positive 63/400.\n\nSo, I think that\u0027s correct. So, the coefficient is 63/400.\n\n**Final Answer**\nThe coefficient of \\( x^2 y^6 \\) is \\boxed{\\dfrac{63}{400}}.\n\u003c/think\u003e\n\nTo find the coefficient of \\( x^2 y^6 \\) in the expansion of \\( \\left( \\frac{3}{5}x - \\frac{y}{2} \\right)^8 \\), we use the binomial theorem. The binomial theorem states that:\n\n\\[\n(a + b)^n = \\sum_{k=0}^{n} \\binom{n}{k} a^{n - k} b^k\n\\]\n\nHere, \\( a = \\frac{3}{5}x \\), \\( b = -\\frac{y}{2} \\), and \\( n = 8 \\). We need the term where the power of \\( x \\) is 2 and the power of \\( y \\) is 6. This corresponds to \\( a^2 b^6 \\).\n\nFirst, we determine the binomial coefficient \\( \\binom{8}{6} \\):\n\n\\[\n\\binom{8}{6} = \\frac{8!}{6! \\cdot 2!} = \\frac{40320}{720 \\cdot 2} = 28\n\\]\n\nNext, we compute \\( a^2 \\):\n\n\\[\na^2 = \\left( \\frac{3}{5}x \\right)^2 = \\left( \\frac{3}{5} \\right)^2 x^2 = \\frac{9}{25}x^2\n\\]\n\nThen, we compute \\( b^6 \\):\n\n\\[\nb^6 = \\left( -\\frac{y}{2} \\right)^6 = \\left( \\frac{1}{2} \\right)^6 y^6 = \\frac{1}{64} y^6\n\\]\n\nNow, we multiply these parts together:\n\n\\[\n\\text{Term} = \\binom{8}{6} \\cdot a^2 \\cdot b^6 = 28 \\cdot \\frac{9}{25} \\cdot \\frac{1}{64}\n\\]\n\nFirst, multiply the constants:\n\n\\[\n28 \\cdot \\frac{9}{25} = \\frac{252}{25}\n\\]\n\nThen, multiply by \\( \\frac{1}{64} \\):\n\n\\[\n\\frac{252}{25} \\cdot \\frac{1}{64} = \\frac{252}{1600}\n\\]\n\nSimplify the fraction:\n\n\\[\n\\frac{252}{1600} = \\frac{63}{400}\n\\]\n\nThus, the coefficient of \\( x^2 y^6 \\) is \\boxed{\\dfrac{63}{400}}.",
        "statistics_text_generation_0": {
            "input_tokens": 77,
            "output_tokens": 2631
        }
    },
    "generation": "\u003cthink\u003e\nAlright, so I have this problem: I need to find the coefficient of \\( x^2 y^6 \\) in the expansion of \\( \\left( \\frac{3}{5}x - \\frac{y}{2} \\right)^8 \\). Hmm, okay. I remember that when expanding binomials, the general term can be found using the binomial theorem. Let me recall what that is.\n\nThe binomial theorem states that \\( (a + b)^n = \\sum_{k=0}^{n} \\binom{n}{k} a^{n - k} b^k \\). So, in this case, my \\( a \\) is \\( \\frac{3}{5}x \\) and my \\( b \\) is \\( -\\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. Let me think about how that relates to the exponents in the binomial theorem. Since the term \\( x^2 y^6 \\) implies that \\( x \\) is raised to the 2nd power and \\( y \\) is raised to the 6th power, that should correspond to \\( a^{2} b^{6} \\) in the expansion.\n\nWait, let me check that: since \\( a = \\frac{3}{5}x \\), \\( a^2 \\) would be \\( \\left( \\frac{3}{5}x \\right)^2 = \\left( \\frac{3}{5} \\right)^2 x^2 \\). Similarly, \\( b = -\\frac{y}{2} \\), so \\( b^6 = \\left( -\\frac{y}{2} \\right)^6 \\). Hmm, okay, that makes sense.\n\nSo, the general term is \\( \\binom{8}{k} a^{8 - k} b^{k} \\). I need the term where the power of \\( x \\) is 2 and the power of \\( y \\) is 6. Since \\( a \\) contributes the \\( x \\) term and \\( b \\) contributes the \\( y \\) term, I can set up the exponents accordingly.\n\nLet me denote the exponent of \\( a \\) as \\( 8 - k \\) and the exponent of \\( b \\) as \\( k \\). Since \\( a \\) is \\( x \\), its exponent \\( 8 - k \\) should be equal to 2, right? So, \\( 8 - k = 2 \\) which implies \\( k = 6 \\). So, the term we\u0027re interested in is when \\( k = 6 \\).\n\nTherefore, the term is \\( \\binom{8}{6} a^{2} b^{6} \\). Let me compute each part step by step.\n\nFirst, compute the binomial coefficient \\( \\binom{8}{6} \\). I remember that \\( \\binom{n}{k} = \\frac{n!}{k!(n - k)!} \\). So, plugging in the values:\n\n\\( \\binom{8}{6} = \\frac{8!}{6! \\cdot (8 - 6)!} = \\frac{8!}{6! \\cdot 2!} \\). Calculating that:\n\n\\( 8! = 40320 \\), \\( 6! = 720 \\), \\( 2! = 2 \\).\n\nSo, \\( \\frac{40320}{720 \\cdot 2} = \\frac{40320}{1440} \\). Let me compute that: 40320 divided by 1440. Hmm, 1440 times 28 is 40320 because 1440 times 10 is 14400, 1440 times 20 is 28800, and 1440 times 8 is 11520; adding 28800 + 11520 gives 40320. So, 1440 times 28 is 40320, so \\( \\frac{40320}{1440} = 28 \\). So, the binomial coefficient is 28.\n\nNext, compute \\( a^2 \\): \\( a = \\frac{3}{5}x \\), so \\( a^2 = \\left( \\frac{3}{5} \\right)^2 x^2 = \\frac{9}{25}x^2 \\).\n\nThen, compute \\( b^6 \\): \\( b = -\\frac{y}{2} \\), so \\( b^6 = \\left( -\\frac{1}{2} \\right)^6 y^6 \\). Wait, hold on, \\( \\left( -\\frac{1}{2} \\right)^6 \\) is the same as \\( \\left( \\frac{1}{2} \\right)^6 \\) because any even power of a negative number is positive. So, \\( \\left( \\frac{1}{2} \\right)^6 = \\frac{1}{64} \\). Therefore, \\( b^6 = \\frac{1}{64} y^6 \\).\n\nNow, multiply all these parts together: the binomial coefficient, \\( a^2 \\), and \\( b^6 \\).\n\nSo, putting it all together:\n\nTerm = \\( \\binom{8}{6} \\cdot a^2 \\cdot b^6 = 28 \\cdot \\frac{9}{25}x^2 \\cdot \\frac{1}{64}y^6 \\).\n\nNow, let\u0027s multiply the constants:\n\nFirst, multiply 28 and \\( \\frac{9}{25} \\):\n\n28 times \\( \\frac{9}{25} \\) is \\( \\frac{28 \\times 9}{25} \\).\n\nCompute 28 times 9: 28 times 10 is 280, minus 28 is 252. So, \\( \\frac{252}{25} \\).\n\nThen, multiply that by \\( \\frac{1}{64} \\):\n\nSo, \\( \\frac{252}{25} \\times \\frac{1}{64} = \\frac{252}{1600} \\).\n\nSimplify \\( \\frac{252}{1600} \\). Let\u0027s see.\n\nDivide numerator and denominator by 4: 252 \u00f7 4 = 63, 1600 \u00f7 4 = 400.\n\nSo, \\( \\frac{63}{400} \\).\n\nIs that reducible? Let\u0027s check. 63 factors are 7 and 9, 400 factors are 2^4 and 5^2. No common factors, so \\( \\frac{63}{400} \\) is the simplified form.\n\nSo, the coefficient is \\( \\frac{63}{400} \\). Therefore, the coefficient of \\( x^2 y^6 \\) is \\( \\frac{63}{400} \\).\n\nWait a second, let me double-check my calculations to make sure I didn\u0027t make a mistake.\n\nSo, binomial coefficient: 8 choose 6 is 28, that\u0027s correct.\n\n\\( a^2 = \\left( \\frac{3}{5} \\right)^2 = \\frac{9}{25} \\), that\u0027s right.\n\n\\( b^6 = \\left( -\\frac{1}{2} \\right)^6 = \\frac{1}{64} \\), correct.\n\nMultiply 28 * 9 = 252, yes, and 252 divided by 25 * 64. Wait, no: 28 * 9 / (25 * 64). Wait, that\u0027s 28 * 9 = 252, and 25 * 64 = 1600, so 252 / 1600. Then, simplifying, divide numerator and denominator by 4: 63 / 400, which is 0.1575.\n\nIs there a simpler way to compute the coefficient? Let me see.\n\nAlternatively, 28 * 9 = 252, 252 / (25 * 64). Since 25 * 64 is 1600, so 252 / 1600. 252 divided by 4 is 63, 1600 divided by 4 is 400. So, yeah, 63/400. That\u0027s correct.\n\nWait, but let me think again about the signs. \\( b = -\\frac{y}{2} \\), so \\( b^6 = \\left( -\\frac{1}{2} \\right)^6 = \\frac{1}{64} \\), because the exponent is even. So, the sign is positive. Therefore, the coefficient is indeed positive 63/400.\n\nSo, I think that\u0027s correct. So, the coefficient is 63/400.\n\n**Final Answer**\nThe coefficient of \\( x^2 y^6 \\) is \\boxed{\\dfrac{63}{400}}.\n\u003c/think\u003e\n\nTo find the coefficient of \\( x^2 y^6 \\) in the expansion of \\( \\left( \\frac{3}{5}x - \\frac{y}{2} \\right)^8 \\), we use the binomial theorem. The binomial theorem states that:\n\n\\[\n(a + b)^n = \\sum_{k=0}^{n} \\binom{n}{k} a^{n - k} b^k\n\\]\n\nHere, \\( a = \\frac{3}{5}x \\), \\( b = -\\frac{y}{2} \\), and \\( n = 8 \\). We need the term where the power of \\( x \\) is 2 and the power of \\( y \\) is 6. This corresponds to \\( a^2 b^6 \\).\n\nFirst, we determine the binomial coefficient \\( \\binom{8}{6} \\):\n\n\\[\n\\binom{8}{6} = \\frac{8!}{6! \\cdot 2!} = \\frac{40320}{720 \\cdot 2} = 28\n\\]\n\nNext, we compute \\( a^2 \\):\n\n\\[\na^2 = \\left( \\frac{3}{5}x \\right)^2 = \\left( \\frac{3}{5} \\right)^2 x^2 = \\frac{9}{25}x^2\n\\]\n\nThen, we compute \\( b^6 \\):\n\n\\[\nb^6 = \\left( -\\frac{y}{2} \\right)^6 = \\left( \\frac{1}{2} \\right)^6 y^6 = \\frac{1}{64} y^6\n\\]\n\nNow, we multiply these parts together:\n\n\\[\n\\text{Term} = \\binom{8}{6} \\cdot a^2 \\cdot b^6 = 28 \\cdot \\frac{9}{25} \\cdot \\frac{1}{64}\n\\]\n\nFirst, multiply the constants:\n\n\\[\n28 \\cdot \\frac{9}{25} = \\frac{252}{25}\n\\]\n\nThen, multiply by \\( \\frac{1}{64} \\):\n\n\\[\n\\frac{252}{25} \\cdot \\frac{1}{64} = \\frac{252}{1600}\n\\]\n\nSimplify the fraction:\n\n\\[\n\\frac{252}{1600} = \\frac{63}{400}\n\\]\n\nThus, the coefficient of \\( x^2 y^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("3482brother/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("3482brother/numina-deepseek-r1-qwen-7b")
```


</details>




