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: 172568
    num_examples: 10
  download_size: 97040
  dataset_size: 172568
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-14b

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/lee1g/numina-deepseek-r1-qwen-14b/raw/main/pipeline.yaml"
```

or explore the configuration:

```console
distilabel pipeline info --config "https://huggingface.co/datasets/lee1g/numina-deepseek-r1-qwen-14b/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 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, let me think about how to approach this.\n\nFirst off, when dealing with expansions like this, especially binomial expansions, I remember that we can use the Binomial Theorem. The Binomial Theorem says that \\( (a + b)^n = \\sum_{k=0}^{n} \\binom{n}{k} a^{n - k} b^k \\). So in this case, our expression is \\( \\left( \\frac{3}{5}x - \\frac{y}{2} \\right)^8 \\), which can be seen as \\( (a + b)^8 \\) where \\( a = \\frac{3}{5}x \\) and \\( b = -\\frac{y}{2} \\).\n\nSo, applying the Binomial Theorem, each term in the expansion will look like \\( \\binom{8}{k} \\left( \\frac{3}{5}x \\right)^{8 - k} \\left( -\\frac{y}{2} \\right)^k \\). I need to find the specific term where the powers of x and y are 2 and 6, respectively. That means we\u0027re looking for \\( x^2 y^6 \\).\n\nLet me break this down.\n\nFirstly, let\u0027s denote:\n- The exponent on \\( a = \\frac{3}{5}x \\) is \\( n - k = 8 - k \\).\n- The exponent on \\( b = -\\frac{y}{2} \\) is \\( k \\).\n\nSince we need an \\( x^2 y^6 \\), that means:\n- For the term involving \\( x \\), it\u0027s raised to the power of 2. So, \\( 8 - k = 2 \\).\n- For the term involving \\( y \\), it\u0027s raised to the power of 6. So, \\( k = 6 \\).\n\nWait, so if I set up those equations:\n1. \\( 8 - k = 2 \\) =\u003e Solving for k: Subtracting both sides by 8? Wait no, that would be negative. Let me think again.\n\nIf \\( 8 - k = 2 \\), then adding k to both sides gives \\( 8 = 2 + k \\). Then subtracting 2 from both sides gives \\( k = 6 \\). So yes, that\u0027s consistent with the y exponent being 6 because \\( b^k \\) would be \\( (-\\frac{y}{2})^6 \\).\n\nGood, so we\u0027re looking for the term where \\( k = 6 \\). That makes sense.\n\nSo now, let me write down the general formula again:\n\n\\( \\binom{8}{k} \\left( \\frac{3}{5}x \\right)^{8 - k} \\left( -\\frac{y}{2} \\right)^k \\).\n\nPlugging in \\( k = 6 \\):\n\nFirst, compute the combination: \\( \\binom{8}{6} \\). I remember that combinations can be calculated as \\( \\binom{n}{k} = \\frac{n!}{k!(n - k)!} \\), so:\n\n\\( \\binom{8}{6} = \\frac{8!}{6!2!} \\).\n\nCalculating this, 8! is 40320, 6! is 720, and 2! is 2. So,\n\n\\( \\frac{40320}{(720)(2)} = \\frac{40320}{1440} \\).\n\nDividing 40320 by 1440: Let\u0027s see, 1440 x 28 is 40320 (because 1440 x 10=14400; 1440x20=28800; 1440x8=11520; adding up: 28800 + 11520 = 40320). So, it\u0027s 28. Therefore, \\( \\binom{8}{6} = 28 \\).\n\nNext, let\u0027s compute the other parts:\n\n\\( \\left( \\frac{3}{5}x \\right)^{8 - 6} = \\left( \\frac{3}{5}x \\right)^2 \\). So that\u0027s \\( \\left(\\frac{3}{5}\\right)^2 x^2 = \\frac{9}{25}x^2 \\).\n\nThen, \\( \\left( -\\frac{y}{2} \\right)^6 \\). Hmm, since the exponent is even, this will make it positive. So,\n\n\\( (-1)^6 \\times \\left(\\frac{y}{2}\\right)^6 = 1 \\times \\frac{y^6}{64} = \\frac{y^6}{64} \\).\n\nAlright, so now putting all these together:\n\nThe term is \\( 28 \\times \\frac{9}{25}x^2 \\times \\frac{y^6}{64} \\).\n\nLet me compute the constants first.\n\nFirst, multiply 28 and 9: 28 x 9. Let\u0027s see, 20x9=180, 8x9=72; adding them gives 252.\n\nThen, divide by 25, so 252 / 25.\n\nNext, multiply that by 1/64: So (252 / 25) x (1 / 64).\n\nWait, let me compute this step-by-step:\n\nFirst, 28 * 9 = 252.\n\nThen, 252 divided by 25 is equal to 10.08? Wait, no, probably better to keep it as a fraction: 252/25.\n\nThen, multiply that by 1/64:\n\nSo, (252/25) * (1/64) = 252 / (25*64).\n\nCompute 25 x 64. Hmm, 25x60=1500 and 25x4=100, so total is 1600.\n\nSo, denominator is 1600.\n\nNumerator is 252.\n\nTherefore, the coefficient is 252 / 1600.\n\nNow, let\u0027s simplify this fraction.\n\nFirst, both numerator and denominator are even numbers, so divide numerator and denominator by 2:\n\n252 \u00f7 2 = 126\n\n1600 \u00f7 2 = 800\n\nSo now, it\u0027s 126 / 800.\n\nAgain, both divisible by 2:\n\n126 \u00f7 2 = 63\n\n800 \u00f7 2 = 400\n\nNow we have 63 / 400.\n\nCheck if this can be reduced further. 63 is 7x9; 400 is 16x25.\n\nNo common factors between numerator and denominator, so 63/400 is the simplified form.\n\nTherefore, the coefficient of \\( x^2 y^6 \\) in the expansion is 63/400.\n\nWait a second. Let me double-check my calculations to make sure I didn\u0027t make any mistakes.\n\nFirst, binomial coefficient: \\( \\binom{8}{6} = 28 \\). That\u0027s correct because it\u0027s equal to \\( \\binom{8}{2} \\), which is also 28.\n\nNext, \\( (3/5)^2 = 9/25. Correct.\n\n\\( (-1/2)^6: Since exponent is even, positive; and (1/2)^6 is 1/64. So that\u0027s correct.\n\nThen, multiplying all constants together:\n\n28 * (9/25) * (1/64).\n\nCompute 28 * 9 = 252.\n\n252 / 25 * 1 / 64 = 252 / (25*64) = 252 / 1600.\n\nDivide numerator and denominator by 2: 126 / 800, then by 2 again: 63/400. Yes, correct.\n\nSo, yes, 63/400 is the coefficient.\n\nTherefore, I think that\u0027s my answer.\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 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 \\):\n\n1. Calculate the binomial coefficient:\n   \\[\n   \\binom{8}{6} = 28\n   \\]\n\n2. Compute \\( \\left( \\frac{3}{5}x \\right)^{2} \\):\n   \\[\n   \\left( \\frac{3}{5} \\right)^2 x^2 = \\frac{9}{25} x^2\n   \\]\n\n3. Compute \\( \\left( -\\frac{y}{2} \\right)^6 \\):\n   \\[\n   (-1)^6 \\left( \\frac{y}{2} \\right)^6 = 1 \\cdot \\frac{y^6}{64}\n   \\]\n\n4. Multiply these results together with the binomial coefficient:\n   \\[\n   28 \\times \\frac{9}{25} x^2 \\times \\frac{y^6}{64}\n   \\]\n\n5. Simplify the constants:\n   \\[\n   28 \\times \\frac{9}{25} \\times \\frac{1}{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": 76,
            "output_tokens": 2364
        }
    },
    "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, let me think about how to approach this.\n\nFirst off, when dealing with expansions like this, especially binomial expansions, I remember that we can use the Binomial Theorem. The Binomial Theorem says that \\( (a + b)^n = \\sum_{k=0}^{n} \\binom{n}{k} a^{n - k} b^k \\). So in this case, our expression is \\( \\left( \\frac{3}{5}x - \\frac{y}{2} \\right)^8 \\), which can be seen as \\( (a + b)^8 \\) where \\( a = \\frac{3}{5}x \\) and \\( b = -\\frac{y}{2} \\).\n\nSo, applying the Binomial Theorem, each term in the expansion will look like \\( \\binom{8}{k} \\left( \\frac{3}{5}x \\right)^{8 - k} \\left( -\\frac{y}{2} \\right)^k \\). I need to find the specific term where the powers of x and y are 2 and 6, respectively. That means we\u0027re looking for \\( x^2 y^6 \\).\n\nLet me break this down.\n\nFirstly, let\u0027s denote:\n- The exponent on \\( a = \\frac{3}{5}x \\) is \\( n - k = 8 - k \\).\n- The exponent on \\( b = -\\frac{y}{2} \\) is \\( k \\).\n\nSince we need an \\( x^2 y^6 \\), that means:\n- For the term involving \\( x \\), it\u0027s raised to the power of 2. So, \\( 8 - k = 2 \\).\n- For the term involving \\( y \\), it\u0027s raised to the power of 6. So, \\( k = 6 \\).\n\nWait, so if I set up those equations:\n1. \\( 8 - k = 2 \\) =\u003e Solving for k: Subtracting both sides by 8? Wait no, that would be negative. Let me think again.\n\nIf \\( 8 - k = 2 \\), then adding k to both sides gives \\( 8 = 2 + k \\). Then subtracting 2 from both sides gives \\( k = 6 \\). So yes, that\u0027s consistent with the y exponent being 6 because \\( b^k \\) would be \\( (-\\frac{y}{2})^6 \\).\n\nGood, so we\u0027re looking for the term where \\( k = 6 \\). That makes sense.\n\nSo now, let me write down the general formula again:\n\n\\( \\binom{8}{k} \\left( \\frac{3}{5}x \\right)^{8 - k} \\left( -\\frac{y}{2} \\right)^k \\).\n\nPlugging in \\( k = 6 \\):\n\nFirst, compute the combination: \\( \\binom{8}{6} \\). I remember that combinations can be calculated as \\( \\binom{n}{k} = \\frac{n!}{k!(n - k)!} \\), so:\n\n\\( \\binom{8}{6} = \\frac{8!}{6!2!} \\).\n\nCalculating this, 8! is 40320, 6! is 720, and 2! is 2. So,\n\n\\( \\frac{40320}{(720)(2)} = \\frac{40320}{1440} \\).\n\nDividing 40320 by 1440: Let\u0027s see, 1440 x 28 is 40320 (because 1440 x 10=14400; 1440x20=28800; 1440x8=11520; adding up: 28800 + 11520 = 40320). So, it\u0027s 28. Therefore, \\( \\binom{8}{6} = 28 \\).\n\nNext, let\u0027s compute the other parts:\n\n\\( \\left( \\frac{3}{5}x \\right)^{8 - 6} = \\left( \\frac{3}{5}x \\right)^2 \\). So that\u0027s \\( \\left(\\frac{3}{5}\\right)^2 x^2 = \\frac{9}{25}x^2 \\).\n\nThen, \\( \\left( -\\frac{y}{2} \\right)^6 \\). Hmm, since the exponent is even, this will make it positive. So,\n\n\\( (-1)^6 \\times \\left(\\frac{y}{2}\\right)^6 = 1 \\times \\frac{y^6}{64} = \\frac{y^6}{64} \\).\n\nAlright, so now putting all these together:\n\nThe term is \\( 28 \\times \\frac{9}{25}x^2 \\times \\frac{y^6}{64} \\).\n\nLet me compute the constants first.\n\nFirst, multiply 28 and 9: 28 x 9. Let\u0027s see, 20x9=180, 8x9=72; adding them gives 252.\n\nThen, divide by 25, so 252 / 25.\n\nNext, multiply that by 1/64: So (252 / 25) x (1 / 64).\n\nWait, let me compute this step-by-step:\n\nFirst, 28 * 9 = 252.\n\nThen, 252 divided by 25 is equal to 10.08? Wait, no, probably better to keep it as a fraction: 252/25.\n\nThen, multiply that by 1/64:\n\nSo, (252/25) * (1/64) = 252 / (25*64).\n\nCompute 25 x 64. Hmm, 25x60=1500 and 25x4=100, so total is 1600.\n\nSo, denominator is 1600.\n\nNumerator is 252.\n\nTherefore, the coefficient is 252 / 1600.\n\nNow, let\u0027s simplify this fraction.\n\nFirst, both numerator and denominator are even numbers, so divide numerator and denominator by 2:\n\n252 \u00f7 2 = 126\n\n1600 \u00f7 2 = 800\n\nSo now, it\u0027s 126 / 800.\n\nAgain, both divisible by 2:\n\n126 \u00f7 2 = 63\n\n800 \u00f7 2 = 400\n\nNow we have 63 / 400.\n\nCheck if this can be reduced further. 63 is 7x9; 400 is 16x25.\n\nNo common factors between numerator and denominator, so 63/400 is the simplified form.\n\nTherefore, the coefficient of \\( x^2 y^6 \\) in the expansion is 63/400.\n\nWait a second. Let me double-check my calculations to make sure I didn\u0027t make any mistakes.\n\nFirst, binomial coefficient: \\( \\binom{8}{6} = 28 \\). That\u0027s correct because it\u0027s equal to \\( \\binom{8}{2} \\), which is also 28.\n\nNext, \\( (3/5)^2 = 9/25. Correct.\n\n\\( (-1/2)^6: Since exponent is even, positive; and (1/2)^6 is 1/64. So that\u0027s correct.\n\nThen, multiplying all constants together:\n\n28 * (9/25) * (1/64).\n\nCompute 28 * 9 = 252.\n\n252 / 25 * 1 / 64 = 252 / (25*64) = 252 / 1600.\n\nDivide numerator and denominator by 2: 126 / 800, then by 2 again: 63/400. Yes, correct.\n\nSo, yes, 63/400 is the coefficient.\n\nTherefore, I think that\u0027s my answer.\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 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 \\):\n\n1. Calculate the binomial coefficient:\n   \\[\n   \\binom{8}{6} = 28\n   \\]\n\n2. Compute \\( \\left( \\frac{3}{5}x \\right)^{2} \\):\n   \\[\n   \\left( \\frac{3}{5} \\right)^2 x^2 = \\frac{9}{25} x^2\n   \\]\n\n3. Compute \\( \\left( -\\frac{y}{2} \\right)^6 \\):\n   \\[\n   (-1)^6 \\left( \\frac{y}{2} \\right)^6 = 1 \\cdot \\frac{y^6}{64}\n   \\]\n\n4. Multiply these results together with the binomial coefficient:\n   \\[\n   28 \\times \\frac{9}{25} x^2 \\times \\frac{y^6}{64}\n   \\]\n\n5. Simplify the constants:\n   \\[\n   28 \\times \\frac{9}{25} \\times \\frac{1}{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-r1:14b",
    "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("lee1g/numina-deepseek-r1-qwen-14b", "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("lee1g/numina-deepseek-r1-qwen-14b")
```


</details>




