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: 851156
    num_examples: 40
  download_size: 333592
  dataset_size: 851156
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/JinnP/numina-deepseek-r1-qwen-7b/raw/main/pipeline.yaml"
```

or explore the configuration:

```console
distilabel pipeline info --config "https://huggingface.co/datasets/JinnP/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 the term \\( x^2 y^6 \\) in the expansion of \\( \\left( \\frac{3}{5}x - \\frac{y}{2} \\right)^8 \\). Hmm, okay. I think this involves the binomial theorem. Let me recall how that works.\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, our \\( a \\) is \\( \\frac{3}{5}x \\) and our \\( b \\) is \\( -\\frac{y}{2} \\), right? And \\( n \\) is 8.\n\nWe are looking for the term where the power of \\( x \\) is 2 and the power of \\( y \\) is 6. That means we need to find the term where \\( a \\) is raised to the power of 2 and \\( b \\) is raised to the power of 6. Wait, because \\( a \\) has \\( x \\) and \\( b \\) has \\( y \\). So, if \\( x \\) is squared, that would correspond to \\( a \\) being squared, which would mean \\( n - k = 2 \\). Since \\( n = 8 \\), that would mean \\( k = 6 \\). So, the term we\u0027re interested in is the one where \\( k = 6 \\).\n\nLet me write that down. The general term in the expansion is \\( \\binom{8}{k} \\left( \\frac{3}{5}x \\right)^{8 - k} \\left( -\\frac{y}{2} \\right)^k \\).\n\nSo, plugging in \\( k = 6 \\), the term becomes \\( \\binom{8}{6} \\left( \\frac{3}{5}x \\right)^{2} \\left( -\\frac{y}{2} \\right)^6 \\).\n\nNow, I need to compute each part step by step.\n\nFirst, calculate the binomial coefficient \\( \\binom{8}{6} \\). I remember that \\( \\binom{n}{k} = \\frac{n!}{k!(n - k)!} \\). So, \\( \\binom{8}{6} = \\frac{8!}{6!2!} \\). Let me compute that.\n\n8! is 40320, 6! is 720, and 2! is 2. So, \\( \\frac{40320}{720 \\times 2} = \\frac{40320}{1440} = 28 \\). So, the binomial coefficient is 28.\n\nNext, let\u0027s compute \\( \\left( \\frac{3}{5}x \\right)^2 \\). That would be \\( \\left( \\frac{3}{5} \\right)^2 x^2 \\). Calculating \\( \\left( \\frac{3}{5} \\right)^2 \\), that\u0027s \\( \\frac{9}{25} \\). So, this part is \\( \\frac{9}{25}x^2 \\).\n\nNow, the other part is \\( \\left( -\\frac{y}{2} \\right)^6 \\). Let\u0027s compute that. Since the exponent is even, the result will be positive. So, \\( \\left( -\\frac{1}{2} \\right)^6 = \\left( \\frac{1}{2} \\right)^6 = \\frac{1}{64} \\). So, putting it all together, this term is \\( \\frac{1}{64} y^6 \\).\n\nNow, multiplying all these together: 28 (from the binomial coefficient) multiplied by \\( \\frac{9}{25} \\) (from \\( a^2 \\)) multiplied by \\( \\frac{1}{64} \\) (from \\( b^6 \\)).\n\nLet me compute that step by step.\n\nFirst, multiply 28 and \\( \\frac{9}{25} \\). 28 times 9 is 252, so that gives \\( \\frac{252}{25} \\).\n\nThen, multiply that by \\( \\frac{1}{64} \\). So, \\( \\frac{252}{25} \\times \\frac{1}{64} = \\frac{252}{1600} \\).\n\nNow, I can simplify \\( \\frac{252}{1600} \\). Let\u0027s see. Both numerator and denominator are divisible by 4. 252 divided by 4 is 63, and 1600 divided by 4 is 400. So, the fraction becomes \\( \\frac{63}{400} \\).\n\nWait, is 63 and 400 divisible by any common factors? 63 is 7 times 9, and 400 is 16 times 25, which are 2^4 and 5^2. So, no, they don\u0027t share any common factors besides 1. So, \\( \\frac{63}{400} \\) is the simplified form.\n\nHold on, let me double-check my calculations to make sure I didn\u0027t make a mistake.\n\nFirst, binomial coefficient: \\( \\binom{8}{6} = 28 \\). That\u0027s correct because \\( \\binom{8}{2} = 28 \\) as well since \\( \\binom{n}{k} = \\binom{n}{n - k} \\).\n\nThen, \\( \\left( \\frac{3}{5} \\right)^2 = \\frac{9}{25} \\). That\u0027s correct.\n\n\\( \\left( -\\frac{1}{2} \\right)^6 = \\frac{1}{64} \\). Correct, because even power makes it positive.\n\nThen, 28 times \\( \\frac{9}{25} \\) is \\( \\frac{252}{25} \\). Correct.\n\nMultiply that by \\( \\frac{1}{64} \\): \\( \\frac{252}{25} \\times \\frac{1}{64} = \\frac{252}{1600} \\). Simplify by dividing numerator and denominator by 4: 252 \u00f7 4 = 63, 1600 \u00f7 4 = 400. So, \\( \\frac{63}{400} \\). That seems right.\n\nWait, but let me check the arithmetic again. 28 times 9: 28*9. 20*9=180, 8*9=72, so total is 180+72=252. So, that\u0027s correct.\n\nThen, 252 divided by 25 is 10.08, but since we\u0027re dealing with fractions, we don\u0027t need to convert to decimal.\n\nThen, 252/25 * 1/64: indeed, 252/25 * 1/64 is 252/(25*64) = 252/1600. Then simplifying by dividing numerator and denominator by 4 gives 63/400.\n\nSo, that seems right.\n\nWait a second, but is the coefficient positive or negative? Let\u0027s think.\n\nIn the term \\( \\left( -\\frac{y}{2} \\right)^6 \\), since the exponent is even, it becomes positive. So, the coefficient is positive. So, 63/400 is correct.\n\nTherefore, the coefficient of \\( x^2 y^6 \\) is \\( \\frac{63}{400} \\).\n\nHmm, that seems correct.\n\nAlternatively, maybe I can think of it another way. Let me compute the entire term:\n\n\\( \\binom{8}{6} \\times \\left( \\frac{3}{5}x \\right)^2 \\times \\left( -\\frac{y}{2} \\right)^6 \\).\n\nWhich is 28 * (9/25 x\u00b2) * (1/64 y^6).\n\nMultiplying the coefficients: 28 * 9/25 * 1/64.\n\n28 * 9 = 252; 25 * 64 = 1600.\n\nSo, 252/1600, simplifies to 63/400 as before.\n\nYes, that\u0027s consistent.\n\nLet me just verify if I applied the binomial theorem correctly.\n\nYes, the binomial theorem is for any term \\( (a + b)^n \\), and each term is \\( \\binom{n}{k} a^{n - k} b^k \\). In our case, \\( a = \\frac{3}{5}x \\), \\( b = -\\frac{y}{2} \\), so each term is \\( \\binom{8}{k} \\left( \\frac{3}{5}x \\right)^{8 - k} \\left( -\\frac{y}{2} \\right)^k \\). We needed \\( x^2 y^6 \\), so \\( 8 - k = 2 \\) which gives \\( k = 6 \\). So, we plug in \\( k = 6 \\) and compute the term. That seems correct.\n\nAlso, note that when you raise \\( -\\frac{y}{2} \\) to the 6th power, since 6 is even, it becomes positive \\( \\left( \\frac{1}{2} \\right)^6 = \\frac{1}{64} \\). So, that\u0027s correct.\n\nSo, I think the coefficient is indeed \\( \\frac{63}{400} \\).\n\nJust to be thorough, let me see if I can write out the term again step by step:\n\n1. Binomial coefficient: \\( \\binom{8}{6} = 28 \\).\n2. \\( \\left( \\frac{3}{5} \\right)^2 = \\frac{9}{25} \\).\n3. \\( \\left( -\\frac{1}{2} \\right)^6 = \\frac{1}{64} \\).\n4. Multiply all together: \\( 28 * \\frac{9}{25} * \\frac{1}{64} \\).\n5. \\( 28 * 9 = 252 \\).\n6. \\( 252 / (25 * 64) = 252 / 1600 \\).\n7. Simplify: divide numerator and denominator by 4: 63/400.\n\nYep, that\u0027s consistent.\n\nSo, I think I\u0027m confident that 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 \\( k = 6 \\) because \\( 8 - k = 2 \\).\n\nThe term is given by:\n\n\\[\n\\binom{8}{6} \\left( \\frac{3}{5}x \\right)^{2} \\left( -\\frac{y}{2} \\right)^6\n\\]\n\nFirst, we calculate the binomial coefficient:\n\n\\[\n\\binom{8}{6} = \\frac{8!}{6!2!} = \\frac{40320}{720 \\times 2} = 28\n\\]\n\nNext, we compute \\( \\left( \\frac{3}{5}x \\right)^2 \\):\n\n\\[\n\\left( \\frac{3}{5} \\right)^2 x^2 = \\frac{9}{25} x^2\n\\]\n\nThen, we compute \\( \\left( -\\frac{y}{2} \\right)^6 \\):\n\n\\[\n\\left( -\\frac{1}{2} \\right)^6 = \\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\nSimplify \\( \\frac{252}{1600} \\) by dividing both the numerator and the denominator by 4:\n\n\\[\n\\frac{252 \\div 4}{1600 \\div 4} = \\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": 3038
        }
    },
    "generation": "\u003cthink\u003e\nAlright, so I need to find the coefficient of the term \\( x^2 y^6 \\) in the expansion of \\( \\left( \\frac{3}{5}x - \\frac{y}{2} \\right)^8 \\). Hmm, okay. I think this involves the binomial theorem. Let me recall how that works.\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, our \\( a \\) is \\( \\frac{3}{5}x \\) and our \\( b \\) is \\( -\\frac{y}{2} \\), right? And \\( n \\) is 8.\n\nWe are looking for the term where the power of \\( x \\) is 2 and the power of \\( y \\) is 6. That means we need to find the term where \\( a \\) is raised to the power of 2 and \\( b \\) is raised to the power of 6. Wait, because \\( a \\) has \\( x \\) and \\( b \\) has \\( y \\). So, if \\( x \\) is squared, that would correspond to \\( a \\) being squared, which would mean \\( n - k = 2 \\). Since \\( n = 8 \\), that would mean \\( k = 6 \\). So, the term we\u0027re interested in is the one where \\( k = 6 \\).\n\nLet me write that down. The general term in the expansion is \\( \\binom{8}{k} \\left( \\frac{3}{5}x \\right)^{8 - k} \\left( -\\frac{y}{2} \\right)^k \\).\n\nSo, plugging in \\( k = 6 \\), the term becomes \\( \\binom{8}{6} \\left( \\frac{3}{5}x \\right)^{2} \\left( -\\frac{y}{2} \\right)^6 \\).\n\nNow, I need to compute each part step by step.\n\nFirst, calculate the binomial coefficient \\( \\binom{8}{6} \\). I remember that \\( \\binom{n}{k} = \\frac{n!}{k!(n - k)!} \\). So, \\( \\binom{8}{6} = \\frac{8!}{6!2!} \\). Let me compute that.\n\n8! is 40320, 6! is 720, and 2! is 2. So, \\( \\frac{40320}{720 \\times 2} = \\frac{40320}{1440} = 28 \\). So, the binomial coefficient is 28.\n\nNext, let\u0027s compute \\( \\left( \\frac{3}{5}x \\right)^2 \\). That would be \\( \\left( \\frac{3}{5} \\right)^2 x^2 \\). Calculating \\( \\left( \\frac{3}{5} \\right)^2 \\), that\u0027s \\( \\frac{9}{25} \\). So, this part is \\( \\frac{9}{25}x^2 \\).\n\nNow, the other part is \\( \\left( -\\frac{y}{2} \\right)^6 \\). Let\u0027s compute that. Since the exponent is even, the result will be positive. So, \\( \\left( -\\frac{1}{2} \\right)^6 = \\left( \\frac{1}{2} \\right)^6 = \\frac{1}{64} \\). So, putting it all together, this term is \\( \\frac{1}{64} y^6 \\).\n\nNow, multiplying all these together: 28 (from the binomial coefficient) multiplied by \\( \\frac{9}{25} \\) (from \\( a^2 \\)) multiplied by \\( \\frac{1}{64} \\) (from \\( b^6 \\)).\n\nLet me compute that step by step.\n\nFirst, multiply 28 and \\( \\frac{9}{25} \\). 28 times 9 is 252, so that gives \\( \\frac{252}{25} \\).\n\nThen, multiply that by \\( \\frac{1}{64} \\). So, \\( \\frac{252}{25} \\times \\frac{1}{64} = \\frac{252}{1600} \\).\n\nNow, I can simplify \\( \\frac{252}{1600} \\). Let\u0027s see. Both numerator and denominator are divisible by 4. 252 divided by 4 is 63, and 1600 divided by 4 is 400. So, the fraction becomes \\( \\frac{63}{400} \\).\n\nWait, is 63 and 400 divisible by any common factors? 63 is 7 times 9, and 400 is 16 times 25, which are 2^4 and 5^2. So, no, they don\u0027t share any common factors besides 1. So, \\( \\frac{63}{400} \\) is the simplified form.\n\nHold on, let me double-check my calculations to make sure I didn\u0027t make a mistake.\n\nFirst, binomial coefficient: \\( \\binom{8}{6} = 28 \\). That\u0027s correct because \\( \\binom{8}{2} = 28 \\) as well since \\( \\binom{n}{k} = \\binom{n}{n - k} \\).\n\nThen, \\( \\left( \\frac{3}{5} \\right)^2 = \\frac{9}{25} \\). That\u0027s correct.\n\n\\( \\left( -\\frac{1}{2} \\right)^6 = \\frac{1}{64} \\). Correct, because even power makes it positive.\n\nThen, 28 times \\( \\frac{9}{25} \\) is \\( \\frac{252}{25} \\). Correct.\n\nMultiply that by \\( \\frac{1}{64} \\): \\( \\frac{252}{25} \\times \\frac{1}{64} = \\frac{252}{1600} \\). Simplify by dividing numerator and denominator by 4: 252 \u00f7 4 = 63, 1600 \u00f7 4 = 400. So, \\( \\frac{63}{400} \\). That seems right.\n\nWait, but let me check the arithmetic again. 28 times 9: 28*9. 20*9=180, 8*9=72, so total is 180+72=252. So, that\u0027s correct.\n\nThen, 252 divided by 25 is 10.08, but since we\u0027re dealing with fractions, we don\u0027t need to convert to decimal.\n\nThen, 252/25 * 1/64: indeed, 252/25 * 1/64 is 252/(25*64) = 252/1600. Then simplifying by dividing numerator and denominator by 4 gives 63/400.\n\nSo, that seems right.\n\nWait a second, but is the coefficient positive or negative? Let\u0027s think.\n\nIn the term \\( \\left( -\\frac{y}{2} \\right)^6 \\), since the exponent is even, it becomes positive. So, the coefficient is positive. So, 63/400 is correct.\n\nTherefore, the coefficient of \\( x^2 y^6 \\) is \\( \\frac{63}{400} \\).\n\nHmm, that seems correct.\n\nAlternatively, maybe I can think of it another way. Let me compute the entire term:\n\n\\( \\binom{8}{6} \\times \\left( \\frac{3}{5}x \\right)^2 \\times \\left( -\\frac{y}{2} \\right)^6 \\).\n\nWhich is 28 * (9/25 x\u00b2) * (1/64 y^6).\n\nMultiplying the coefficients: 28 * 9/25 * 1/64.\n\n28 * 9 = 252; 25 * 64 = 1600.\n\nSo, 252/1600, simplifies to 63/400 as before.\n\nYes, that\u0027s consistent.\n\nLet me just verify if I applied the binomial theorem correctly.\n\nYes, the binomial theorem is for any term \\( (a + b)^n \\), and each term is \\( \\binom{n}{k} a^{n - k} b^k \\). In our case, \\( a = \\frac{3}{5}x \\), \\( b = -\\frac{y}{2} \\), so each term is \\( \\binom{8}{k} \\left( \\frac{3}{5}x \\right)^{8 - k} \\left( -\\frac{y}{2} \\right)^k \\). We needed \\( x^2 y^6 \\), so \\( 8 - k = 2 \\) which gives \\( k = 6 \\). So, we plug in \\( k = 6 \\) and compute the term. That seems correct.\n\nAlso, note that when you raise \\( -\\frac{y}{2} \\) to the 6th power, since 6 is even, it becomes positive \\( \\left( \\frac{1}{2} \\right)^6 = \\frac{1}{64} \\). So, that\u0027s correct.\n\nSo, I think the coefficient is indeed \\( \\frac{63}{400} \\).\n\nJust to be thorough, let me see if I can write out the term again step by step:\n\n1. Binomial coefficient: \\( \\binom{8}{6} = 28 \\).\n2. \\( \\left( \\frac{3}{5} \\right)^2 = \\frac{9}{25} \\).\n3. \\( \\left( -\\frac{1}{2} \\right)^6 = \\frac{1}{64} \\).\n4. Multiply all together: \\( 28 * \\frac{9}{25} * \\frac{1}{64} \\).\n5. \\( 28 * 9 = 252 \\).\n6. \\( 252 / (25 * 64) = 252 / 1600 \\).\n7. Simplify: divide numerator and denominator by 4: 63/400.\n\nYep, that\u0027s consistent.\n\nSo, I think I\u0027m confident that 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 \\( k = 6 \\) because \\( 8 - k = 2 \\).\n\nThe term is given by:\n\n\\[\n\\binom{8}{6} \\left( \\frac{3}{5}x \\right)^{2} \\left( -\\frac{y}{2} \\right)^6\n\\]\n\nFirst, we calculate the binomial coefficient:\n\n\\[\n\\binom{8}{6} = \\frac{8!}{6!2!} = \\frac{40320}{720 \\times 2} = 28\n\\]\n\nNext, we compute \\( \\left( \\frac{3}{5}x \\right)^2 \\):\n\n\\[\n\\left( \\frac{3}{5} \\right)^2 x^2 = \\frac{9}{25} x^2\n\\]\n\nThen, we compute \\( \\left( -\\frac{y}{2} \\right)^6 \\):\n\n\\[\n\\left( -\\frac{1}{2} \\right)^6 = \\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\nSimplify \\( \\frac{252}{1600} \\) by dividing both the numerator and the denominator by 4:\n\n\\[\n\\frac{252 \\div 4}{1600 \\div 4} = \\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("JinnP/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("JinnP/numina-deepseek-r1-qwen-7b")
```


</details>




