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: 87723
    num_examples: 4
  download_size: 53472
  dataset_size: 87723
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/xiazi-yu/numina-deepseek-r1-qwen-7b/raw/main/pipeline.yaml"
```

or explore the configuration:

```console
distilabel pipeline info --config "https://huggingface.co/datasets/xiazi-yu/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": "Alright, so I have this problem here: I need to find the coefficient of the term \\(x^2y^6\\) in the expansion of \\(\\left(\\frac{3}{5}x - \\frac{y}{2}\\right)^8\\). Hmm, okay. I remember that when you have an expression raised to a power like that, you can use the binomial theorem to expand it. Let me recall how that works.\n\nThe binomial theorem says 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\\) is 8. So, the expansion will have terms of the form \\(\\binom{8}{k} \\left(\\frac{3}{5}x\\right)^{8 - k} \\left(-\\frac{y}{2}\\right)^k\\).\n\nBut the question is specifically asking for the coefficient of \\(x^2y^6\\). So, I need to figure out which term in the expansion will result in \\(x^2y^6\\). Let me think about that. \n\nEach term in the expansion is a product of \\(\\binom{8}{k}\\), \\(\\left(\\frac{3}{5}x\\right)^{8 - k}\\), and \\(\\left(-\\frac{y}{2}\\right)^k\\). So, in each term, the exponent of \\(x\\) is \\(8 - k\\) and the exponent of \\(y\\) is \\(k\\). Therefore, to get \\(x^2y^6\\), we need \\(8 - k = 2\\) for the exponent of \\(x\\), which implies that \\(k = 6\\). \n\nSo, the term corresponding to \\(k = 6\\) will be the one with \\(x^2y^6\\). Let me write that out:\n\nThe term is \\(\\binom{8}{6} \\left(\\frac{3}{5}x\\right)^{8 - 6} \\left(-\\frac{y}{2}\\right)^6\\).\n\nSimplify that:\n\nFirst, compute the binomial coefficient: \\(\\binom{8}{6}\\). I remember that \\(\\binom{n}{k} = \\frac{n!}{k!(n - k)!}\\), so that would be \\(\\frac{8!}{6!2!}\\). Calculating that, 8! is 40320, 6! is 720, and 2! is 2. So, 40320 divided by (720 times 2) is 40320 / 1440. Let me compute that: 40320 / 1440. Hmm, 1440 times 28 is 40320 because 1440*20 is 28800, and 1440*8 is 11520, so 28800 + 11520 is 40320. So, \\(\\binom{8}{6} = 28\\). Got that.\n\nNext, \\(\\left(\\frac{3}{5}x\\right)^{2}\\), since \\(8 - 6 = 2\\). Let me compute that: \\(\\left(\\frac{3}{5}\\right)^2 = \\frac{9}{25}\\), and the exponent on \\(x\\) is 2, so that term is \\(\\frac{9}{25}x^2\\).\n\nNow, the other part is \\(\\left(-\\frac{y}{2}\\right)^6\\). Let\u0027s compute that. First, \\(\\left(-\\frac{1}{2}\\right)^6\\) is \\( \\frac{1}{64}\\), since any negative number raised to an even power is positive. So, that term is \\(\\frac{1}{64}y^6\\).\n\nPutting it all together, the term is:\n\n\\(28 \\times \\frac{9}{25}x^2 \\times \\frac{1}{64}y^6\\).\n\nNow, let\u0027s compute the numerical coefficient. So, 28 multiplied by \\(\\frac{9}{25}\\) multiplied by \\(\\frac{1}{64}\\). Let me write that as:\n\n\\(28 \\times \\frac{9}{25} \\times \\frac{1}{64}\\).\n\nI can simplify this step by step. Maybe first multiply 28 and 9, then divide by 25 and 64.\n\n28 times 9 is 252. So, now we have \\(\\frac{252}{25} \\times \\frac{1}{64}\\). Which is \\(\\frac{252}{25 \\times 64}\\). Let me compute 25 times 64. 25 times 60 is 1500, and 25 times 4 is 100, so 1500 + 100 is 1600. So, that\u0027s \\(\\frac{252}{1600}\\).\n\nSimplify \\(\\frac{252}{1600}\\). Let\u0027s see, both numbers are divisible by 4. 252 divided by 4 is 63, and 1600 divided by 4 is 400. So, that\u0027s \\(\\frac{63}{400}\\). Check if 63 and 400 have any common factors. 63 is 7 times 9, and 400 is 16 times 25. No common factors besides 1, so that\u0027s the simplified fraction.\n\nSo, the coefficient is \\(\\frac{63}{400}\\). Therefore, the term is \\(\\frac{63}{400}x^2y^6\\). So, the coefficient is \\(\\frac{63}{400}\\).\n\nWait, let me double-check in case I made a mistake somewhere. So, binomial coefficient is 28, correct? Yes, because \\(\\binom{8}{6} = 28\\). Then, \\(\\left(\\frac{3}{5}\\right)^2\\) is indeed \\(\\frac{9}{25}\\), and \\(\\left(-\\frac{1}{2}\\right)^6\\) is \\(\\frac{1}{64}\\), since negative to the 6th power is positive. Multiplying all together: 28 times \\(\\frac{9}{25}\\) is \\(\\frac{252}{25}\\), then times \\(\\frac{1}{64}\\) gives \\(\\frac{252}{1600}\\), which simplifies to \\(\\frac{63}{400}\\). That all seems correct.\n\nI guess another way to think about it is using exponents. The exponent on \\(x\\) is 2, which requires that \\(8 - k = 2\\), so \\(k = 6\\). For \\(y\\), the exponent is \\(k\\), so that\u0027s 6, which is also consistent. So, yeah, that term is indeed the one we\u0027re looking for.\n\nIs there another way to cross-verify? Maybe using the general term formula. The term is \\(\\binom{n}{k} a^{n - k} b^k\\), so substituting, we get \\(\\binom{8}{6} \\left(\\frac{3}{5}x\\right)^{2} \\left(-\\frac{y}{2}\\right)^6\\). Yep, that\u0027s exactly what I did earlier.\n\nAnd then, calculating each part. So, 28 times 9/25 is 252/25, then times 1/64 is 252/(25*64). 25*64 is 1600, so 252/1600 simplifies to 63/400. Perfect.\n\nWait, just to make sure, maybe I should compute the entire term step by step without simplifying in between. Let me try:\n\nFirst, compute \\(\\left(\\frac{3}{5}\\right)^2 = \\frac{9}{25}\\), then \\(\\left(-\\frac{1}{2}\\right)^6 = \\frac{1}{64}\\). Multiply those together: \\(\\frac{9}{25} \\times \\frac{1}{64} = \\frac{9}{1600}\\). Then, multiply by the binomial coefficient 28: \\(28 \\times \\frac{9}{1600}\\). 28 times 9 is 252. So, 252 divided by 1600 is indeed 63/400. So, same result. That\u0027s reassuring.\n\nI think I\u0027m confident now that the coefficient is \\(\\frac{63}{400}\\). So, the answer is \\(\\frac{63}{400}\\).\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 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 exponent of \\(x\\) is 2 and the exponent of \\(y\\) is 6. This occurs when \\(8 - k = 2\\) and \\(k = 6\\).\n\nThe term corresponding to \\(k = 6\\) is:\n\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\\[\n\\binom{8}{6} = \\frac{8!}{6!2!} = \\frac{40320}{720 \\times 2} = 28\n\\]\n\nNext, compute \\(\\left(\\frac{3}{5}x\\right)^2\\):\n\n\\[\n\\left(\\frac{3}{5}\\right)^2 = \\frac{9}{25}\n\\]\n\nThen, compute \\(\\left(-\\frac{y}{2}\\right)^6\\):\n\n\\[\n\\left(-\\frac{1}{2}\\right)^6 = \\frac{1}{64}\n\\]\n\nNow, multiply these results together with the binomial coefficient:\n\n\\[\n28 \\times \\frac{9}{25} \\times \\frac{1}{64} = 28 \\times \\frac{9}{1600} = \\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": 78,
            "output_tokens": 2532
        }
    },
    "generation": "Alright, so I have this problem here: I need to find the coefficient of the term \\(x^2y^6\\) in the expansion of \\(\\left(\\frac{3}{5}x - \\frac{y}{2}\\right)^8\\). Hmm, okay. I remember that when you have an expression raised to a power like that, you can use the binomial theorem to expand it. Let me recall how that works.\n\nThe binomial theorem says 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\\) is 8. So, the expansion will have terms of the form \\(\\binom{8}{k} \\left(\\frac{3}{5}x\\right)^{8 - k} \\left(-\\frac{y}{2}\\right)^k\\).\n\nBut the question is specifically asking for the coefficient of \\(x^2y^6\\). So, I need to figure out which term in the expansion will result in \\(x^2y^6\\). Let me think about that. \n\nEach term in the expansion is a product of \\(\\binom{8}{k}\\), \\(\\left(\\frac{3}{5}x\\right)^{8 - k}\\), and \\(\\left(-\\frac{y}{2}\\right)^k\\). So, in each term, the exponent of \\(x\\) is \\(8 - k\\) and the exponent of \\(y\\) is \\(k\\). Therefore, to get \\(x^2y^6\\), we need \\(8 - k = 2\\) for the exponent of \\(x\\), which implies that \\(k = 6\\). \n\nSo, the term corresponding to \\(k = 6\\) will be the one with \\(x^2y^6\\). Let me write that out:\n\nThe term is \\(\\binom{8}{6} \\left(\\frac{3}{5}x\\right)^{8 - 6} \\left(-\\frac{y}{2}\\right)^6\\).\n\nSimplify that:\n\nFirst, compute the binomial coefficient: \\(\\binom{8}{6}\\). I remember that \\(\\binom{n}{k} = \\frac{n!}{k!(n - k)!}\\), so that would be \\(\\frac{8!}{6!2!}\\). Calculating that, 8! is 40320, 6! is 720, and 2! is 2. So, 40320 divided by (720 times 2) is 40320 / 1440. Let me compute that: 40320 / 1440. Hmm, 1440 times 28 is 40320 because 1440*20 is 28800, and 1440*8 is 11520, so 28800 + 11520 is 40320. So, \\(\\binom{8}{6} = 28\\). Got that.\n\nNext, \\(\\left(\\frac{3}{5}x\\right)^{2}\\), since \\(8 - 6 = 2\\). Let me compute that: \\(\\left(\\frac{3}{5}\\right)^2 = \\frac{9}{25}\\), and the exponent on \\(x\\) is 2, so that term is \\(\\frac{9}{25}x^2\\).\n\nNow, the other part is \\(\\left(-\\frac{y}{2}\\right)^6\\). Let\u0027s compute that. First, \\(\\left(-\\frac{1}{2}\\right)^6\\) is \\( \\frac{1}{64}\\), since any negative number raised to an even power is positive. So, that term is \\(\\frac{1}{64}y^6\\).\n\nPutting it all together, the term is:\n\n\\(28 \\times \\frac{9}{25}x^2 \\times \\frac{1}{64}y^6\\).\n\nNow, let\u0027s compute the numerical coefficient. So, 28 multiplied by \\(\\frac{9}{25}\\) multiplied by \\(\\frac{1}{64}\\). Let me write that as:\n\n\\(28 \\times \\frac{9}{25} \\times \\frac{1}{64}\\).\n\nI can simplify this step by step. Maybe first multiply 28 and 9, then divide by 25 and 64.\n\n28 times 9 is 252. So, now we have \\(\\frac{252}{25} \\times \\frac{1}{64}\\). Which is \\(\\frac{252}{25 \\times 64}\\). Let me compute 25 times 64. 25 times 60 is 1500, and 25 times 4 is 100, so 1500 + 100 is 1600. So, that\u0027s \\(\\frac{252}{1600}\\).\n\nSimplify \\(\\frac{252}{1600}\\). Let\u0027s see, both numbers are divisible by 4. 252 divided by 4 is 63, and 1600 divided by 4 is 400. So, that\u0027s \\(\\frac{63}{400}\\). Check if 63 and 400 have any common factors. 63 is 7 times 9, and 400 is 16 times 25. No common factors besides 1, so that\u0027s the simplified fraction.\n\nSo, the coefficient is \\(\\frac{63}{400}\\). Therefore, the term is \\(\\frac{63}{400}x^2y^6\\). So, the coefficient is \\(\\frac{63}{400}\\).\n\nWait, let me double-check in case I made a mistake somewhere. So, binomial coefficient is 28, correct? Yes, because \\(\\binom{8}{6} = 28\\). Then, \\(\\left(\\frac{3}{5}\\right)^2\\) is indeed \\(\\frac{9}{25}\\), and \\(\\left(-\\frac{1}{2}\\right)^6\\) is \\(\\frac{1}{64}\\), since negative to the 6th power is positive. Multiplying all together: 28 times \\(\\frac{9}{25}\\) is \\(\\frac{252}{25}\\), then times \\(\\frac{1}{64}\\) gives \\(\\frac{252}{1600}\\), which simplifies to \\(\\frac{63}{400}\\). That all seems correct.\n\nI guess another way to think about it is using exponents. The exponent on \\(x\\) is 2, which requires that \\(8 - k = 2\\), so \\(k = 6\\). For \\(y\\), the exponent is \\(k\\), so that\u0027s 6, which is also consistent. So, yeah, that term is indeed the one we\u0027re looking for.\n\nIs there another way to cross-verify? Maybe using the general term formula. The term is \\(\\binom{n}{k} a^{n - k} b^k\\), so substituting, we get \\(\\binom{8}{6} \\left(\\frac{3}{5}x\\right)^{2} \\left(-\\frac{y}{2}\\right)^6\\). Yep, that\u0027s exactly what I did earlier.\n\nAnd then, calculating each part. So, 28 times 9/25 is 252/25, then times 1/64 is 252/(25*64). 25*64 is 1600, so 252/1600 simplifies to 63/400. Perfect.\n\nWait, just to make sure, maybe I should compute the entire term step by step without simplifying in between. Let me try:\n\nFirst, compute \\(\\left(\\frac{3}{5}\\right)^2 = \\frac{9}{25}\\), then \\(\\left(-\\frac{1}{2}\\right)^6 = \\frac{1}{64}\\). Multiply those together: \\(\\frac{9}{25} \\times \\frac{1}{64} = \\frac{9}{1600}\\). Then, multiply by the binomial coefficient 28: \\(28 \\times \\frac{9}{1600}\\). 28 times 9 is 252. So, 252 divided by 1600 is indeed 63/400. So, same result. That\u0027s reassuring.\n\nI think I\u0027m confident now that the coefficient is \\(\\frac{63}{400}\\). So, the answer is \\(\\frac{63}{400}\\).\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 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 exponent of \\(x\\) is 2 and the exponent of \\(y\\) is 6. This occurs when \\(8 - k = 2\\) and \\(k = 6\\).\n\nThe term corresponding to \\(k = 6\\) is:\n\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\\[\n\\binom{8}{6} = \\frac{8!}{6!2!} = \\frac{40320}{720 \\times 2} = 28\n\\]\n\nNext, compute \\(\\left(\\frac{3}{5}x\\right)^2\\):\n\n\\[\n\\left(\\frac{3}{5}\\right)^2 = \\frac{9}{25}\n\\]\n\nThen, compute \\(\\left(-\\frac{y}{2}\\right)^6\\):\n\n\\[\n\\left(-\\frac{1}{2}\\right)^6 = \\frac{1}{64}\n\\]\n\nNow, multiply these results together with the binomial coefficient:\n\n\\[\n28 \\times \\frac{9}{25} \\times \\frac{1}{64} = 28 \\times \\frac{9}{1600} = \\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": "Distill-Qwen-1.5B",
    "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("xiazi-yu/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("xiazi-yu/numina-deepseek-r1-qwen-7b")
```


</details>




