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: 1961948
    num_examples: 100
  download_size: 678480
  dataset_size: 1961948
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/Rickxz06/numina-deepseek-r1-qwen-7b/raw/main/pipeline.yaml"
```

or explore the configuration:

```console
distilabel pipeline info --config "https://huggingface.co/datasets/Rickxz06/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\nOkay, so I have this problem here: 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, alright, let me try to figure this out step by step.\n\nFirst, I remember that when you have a binomial raised to a power, like \\( (a + b)^n \\), you can use the binomial theorem to expand it. The binomial theorem states that each term in the expansion is of the form \\( \\binom{n}{k} a^{n - k} b^k \\), where \\( \\binom{n}{k} \\) is the binomial coefficient, which can be calculated using combinations.\n\nIn this case, my expression is \\( \\left( \\frac{3}{5}x - \\frac{y}{2} \\right)^8 \\). So, comparing it to \\( (a + b)^n \\), my \\( a \\) is \\( \\frac{3}{5}x \\) and my \\( b \\) is \\( -\\frac{y}{2} \\). The exponent \\( n \\) is 8.\n\nI need the coefficient for the term with \\( x^2 y^6 \\). So, let\u0027s figure out what \\( k \\) should be for that term. Since \\( a \\) is associated with \\( x \\) and \\( b \\) is associated with \\( y \\), the term \\( x^2 y^6 \\) must come from when \\( a \\) is raised to the 2nd power and \\( b \\) is raised to the 6th power.\n\nSo, in the general term from the binomial expansion, \\( \\binom{8}{k} \\left( \\frac{3}{5}x \\right)^{8 - k} \\left( -\\frac{y}{2} \\right)^k \\). I need \\( (8 - k) \\) to be 2 and \\( k \\) to be 6 because \\( x \\) is squared and \\( y \\) is raised to the 6th power. Let me check that: If \\( (8 - k) = 2 \\), then \\( k = 6 \\). Perfect, that makes sense.\n\nSo, plugging \\( k = 6 \\) into the general term:\n\nFirst, the binomial coefficient is \\( \\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\n\\( 8! = 40320 \\), \\( 6! = 720 \\), and \\( 2! = 2 \\). So, \\( \\frac{40320}{720 \\times 2} = \\frac{40320}{1440} \\). Dividing 40320 by 1440: Well, 1440 goes into 40320 exactly 28 times because 1440 x 28 = 40320. So, \\( \\binom{8}{6} = 28 \\).\n\nNext, I need to compute \\( \\left( \\frac{3}{5}x \\right)^{2} \\). Let\u0027s compute the coefficient part: \\( \\left( \\frac{3}{5} \\right)^2 = \\frac{9}{25} \\). So, that gives us \\( \\frac{9}{25}x^2 \\).\n\nThen, the other part is \\( \\left( -\\frac{y}{2} \\right)^6 \\). Hmm, so let\u0027s compute this. First, \\( \\left( -\\frac{1}{2} \\right)^6 \\). Since the exponent is even, the negative sign will become positive. So, \\( \\left( \\frac{1}{2} \\right)^6 = \\frac{1}{64} \\). Therefore, \\( \\left( -\\frac{y}{2} \\right)^6 = \\frac{1}{64}y^6 \\).\n\nNow, putting it all together, the term is:\n\n\\( \\binom{8}{6} \\times \\left( \\frac{9}{25}x^2 \\right) \\times \\left( \\frac{1}{64}y^6 \\right) \\).\n\nMultiplying the coefficients together: \\( 28 \\times \\frac{9}{25} \\times \\frac{1}{64} \\).\n\nLet me compute that step by step. First, multiply 28 and \\( \\frac{9}{25} \\):\n\n\\( 28 \\times \\frac{9}{25} = \\frac{28 \\times 9}{25} \\). 28 times 9 is 252, so that\u0027s \\( \\frac{252}{25} \\).\n\nNext, multiply \\( \\frac{252}{25} \\) by \\( \\frac{1}{64} \\):\n\n\\( \\frac{252}{25} \\times \\frac{1}{64} = \\frac{252}{1600} \\).\n\nNow, let me simplify \\( \\frac{252}{1600} \\). Both numerator and denominator can be divided by 4:\n\n252 \u00f7 4 = 63\n\n1600 \u00f7 4 = 400\n\nSo, \\( \\frac{63}{400} \\). Hmm, can this be simplified further? Let me check: 63 and 400. 63 factors are 7 and 9, 400 is 2^4 and 5^2. So, no common factors, so \\( \\frac{63}{400} \\) is in simplest form.\n\nTherefore, the coefficient is \\( \\frac{63}{400} \\).\n\nWait, let me double-check my calculations to make sure I didn\u0027t make a mistake.\n\nFirst, binomial coefficient: \\( \\binom{8}{6} = 28 \\). That seems correct.\n\n\\( \\left( \\frac{3}{5} \\right)^2 = \\frac{9}{25} \\). That\u0027s right.\n\n\\( \\left( -\\frac{1}{2} \\right)^6 = \\frac{1}{64} \\). Correct, since even power removes the negative.\n\nMultiplying coefficients: 28 times (9/25) times (1/64). Let me compute 28 * 9 first.\n\n28 * 9: 20*9=180, 8*9=72, so 180+72=252. So, 252/25 * 1/64.\n\nMultiplying 252 and 1 gives 252, and 25*64 is 1600. So, 252/1600 reduces to 63/400. Yep, that\u0027s correct.\n\nSo, I think that\u0027s the right coefficient.\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 each term in the expansion is of the form \\( \\binom{n}{k} a^{n - k} b^k \\).\n\nHere, \\( a = \\frac{3}{5}x \\), \\( b = -\\frac{y}{2} \\), and \\( n = 8 \\). We need the term where \\( x \\) is squared and \\( y \\) is raised to the 6th power, which corresponds to \\( k = 6 \\).\n\n1. Calculate the binomial coefficient \\( \\binom{8}{6} \\):\n   \\[\n   \\binom{8}{6} = \\frac{8!}{6!2!} = \\frac{40320}{720 \\times 2} = 28\n   \\]\n\n2. Calculate \\( \\left( \\frac{3}{5}x \\right)^2 \\):\n   \\[\n   \\left( \\frac{3}{5} \\right)^2 = \\frac{9}{25}\n   \\]\n\n3. Calculate \\( \\left( -\\frac{y}{2} \\right)^6 \\):\n   \\[\n   \\left( -\\frac{1}{2} \\right)^6 = \\left( \\frac{1}{2} \\right)^6 = \\frac{1}{64}\n   \\]\n\n4. Multiply the coefficients together:\n   \\[\n   28 \\times \\frac{9}{25} \\times \\frac{1}{64} = \\frac{28 \\times 9}{25 \\times 64} = \\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": 2073
        }
    },
    "generation": "\u003cthink\u003e\nOkay, so I have this problem here: 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, alright, let me try to figure this out step by step.\n\nFirst, I remember that when you have a binomial raised to a power, like \\( (a + b)^n \\), you can use the binomial theorem to expand it. The binomial theorem states that each term in the expansion is of the form \\( \\binom{n}{k} a^{n - k} b^k \\), where \\( \\binom{n}{k} \\) is the binomial coefficient, which can be calculated using combinations.\n\nIn this case, my expression is \\( \\left( \\frac{3}{5}x - \\frac{y}{2} \\right)^8 \\). So, comparing it to \\( (a + b)^n \\), my \\( a \\) is \\( \\frac{3}{5}x \\) and my \\( b \\) is \\( -\\frac{y}{2} \\). The exponent \\( n \\) is 8.\n\nI need the coefficient for the term with \\( x^2 y^6 \\). So, let\u0027s figure out what \\( k \\) should be for that term. Since \\( a \\) is associated with \\( x \\) and \\( b \\) is associated with \\( y \\), the term \\( x^2 y^6 \\) must come from when \\( a \\) is raised to the 2nd power and \\( b \\) is raised to the 6th power.\n\nSo, in the general term from the binomial expansion, \\( \\binom{8}{k} \\left( \\frac{3}{5}x \\right)^{8 - k} \\left( -\\frac{y}{2} \\right)^k \\). I need \\( (8 - k) \\) to be 2 and \\( k \\) to be 6 because \\( x \\) is squared and \\( y \\) is raised to the 6th power. Let me check that: If \\( (8 - k) = 2 \\), then \\( k = 6 \\). Perfect, that makes sense.\n\nSo, plugging \\( k = 6 \\) into the general term:\n\nFirst, the binomial coefficient is \\( \\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\n\\( 8! = 40320 \\), \\( 6! = 720 \\), and \\( 2! = 2 \\). So, \\( \\frac{40320}{720 \\times 2} = \\frac{40320}{1440} \\). Dividing 40320 by 1440: Well, 1440 goes into 40320 exactly 28 times because 1440 x 28 = 40320. So, \\( \\binom{8}{6} = 28 \\).\n\nNext, I need to compute \\( \\left( \\frac{3}{5}x \\right)^{2} \\). Let\u0027s compute the coefficient part: \\( \\left( \\frac{3}{5} \\right)^2 = \\frac{9}{25} \\). So, that gives us \\( \\frac{9}{25}x^2 \\).\n\nThen, the other part is \\( \\left( -\\frac{y}{2} \\right)^6 \\). Hmm, so let\u0027s compute this. First, \\( \\left( -\\frac{1}{2} \\right)^6 \\). Since the exponent is even, the negative sign will become positive. So, \\( \\left( \\frac{1}{2} \\right)^6 = \\frac{1}{64} \\). Therefore, \\( \\left( -\\frac{y}{2} \\right)^6 = \\frac{1}{64}y^6 \\).\n\nNow, putting it all together, the term is:\n\n\\( \\binom{8}{6} \\times \\left( \\frac{9}{25}x^2 \\right) \\times \\left( \\frac{1}{64}y^6 \\right) \\).\n\nMultiplying the coefficients together: \\( 28 \\times \\frac{9}{25} \\times \\frac{1}{64} \\).\n\nLet me compute that step by step. First, multiply 28 and \\( \\frac{9}{25} \\):\n\n\\( 28 \\times \\frac{9}{25} = \\frac{28 \\times 9}{25} \\). 28 times 9 is 252, so that\u0027s \\( \\frac{252}{25} \\).\n\nNext, multiply \\( \\frac{252}{25} \\) by \\( \\frac{1}{64} \\):\n\n\\( \\frac{252}{25} \\times \\frac{1}{64} = \\frac{252}{1600} \\).\n\nNow, let me simplify \\( \\frac{252}{1600} \\). Both numerator and denominator can be divided by 4:\n\n252 \u00f7 4 = 63\n\n1600 \u00f7 4 = 400\n\nSo, \\( \\frac{63}{400} \\). Hmm, can this be simplified further? Let me check: 63 and 400. 63 factors are 7 and 9, 400 is 2^4 and 5^2. So, no common factors, so \\( \\frac{63}{400} \\) is in simplest form.\n\nTherefore, the coefficient is \\( \\frac{63}{400} \\).\n\nWait, let me double-check my calculations to make sure I didn\u0027t make a mistake.\n\nFirst, binomial coefficient: \\( \\binom{8}{6} = 28 \\). That seems correct.\n\n\\( \\left( \\frac{3}{5} \\right)^2 = \\frac{9}{25} \\). That\u0027s right.\n\n\\( \\left( -\\frac{1}{2} \\right)^6 = \\frac{1}{64} \\). Correct, since even power removes the negative.\n\nMultiplying coefficients: 28 times (9/25) times (1/64). Let me compute 28 * 9 first.\n\n28 * 9: 20*9=180, 8*9=72, so 180+72=252. So, 252/25 * 1/64.\n\nMultiplying 252 and 1 gives 252, and 25*64 is 1600. So, 252/1600 reduces to 63/400. Yep, that\u0027s correct.\n\nSo, I think that\u0027s the right coefficient.\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 each term in the expansion is of the form \\( \\binom{n}{k} a^{n - k} b^k \\).\n\nHere, \\( a = \\frac{3}{5}x \\), \\( b = -\\frac{y}{2} \\), and \\( n = 8 \\). We need the term where \\( x \\) is squared and \\( y \\) is raised to the 6th power, which corresponds to \\( k = 6 \\).\n\n1. Calculate the binomial coefficient \\( \\binom{8}{6} \\):\n   \\[\n   \\binom{8}{6} = \\frac{8!}{6!2!} = \\frac{40320}{720 \\times 2} = 28\n   \\]\n\n2. Calculate \\( \\left( \\frac{3}{5}x \\right)^2 \\):\n   \\[\n   \\left( \\frac{3}{5} \\right)^2 = \\frac{9}{25}\n   \\]\n\n3. Calculate \\( \\left( -\\frac{y}{2} \\right)^6 \\):\n   \\[\n   \\left( -\\frac{1}{2} \\right)^6 = \\left( \\frac{1}{2} \\right)^6 = \\frac{1}{64}\n   \\]\n\n4. Multiply the coefficients together:\n   \\[\n   28 \\times \\frac{9}{25} \\times \\frac{1}{64} = \\frac{28 \\times 9}{25 \\times 64} = \\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("Rickxz06/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("Rickxz06/numina-deepseek-r1-qwen-7b")
```


</details>




