{"status": "success", "data": {"description_md": "It's Christmas! As he is financially challenged, Bob can only afford to hang $12$ Christmas lights, $6$ of which are red and $6$ of which are green. He randomly attaches his lights to a string and places them between his house and his neighbor's house. Bob has a *merry Christmas* if in his string of $6$ lights (the ones closer to his house), the red lights are all adjacent and the green lights are all adjacent. The probability that Bob has a merry Christmas can be written as $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.", "description_html": "<p>It&#8217;s Christmas! As he is financially challenged, Bob can only afford to hang <span class=\"katex--inline\">12</span> Christmas lights, <span class=\"katex--inline\">6</span> of which are red and <span class=\"katex--inline\">6</span> of which are green. He randomly attaches his lights to a string and places them between his house and his neighbor&#8217;s house. Bob has a <em>merry Christmas</em> if in his string of <span class=\"katex--inline\">6</span> lights (the ones closer to his house), the red lights are all adjacent and the green lights are all adjacent. The probability that Bob has a merry Christmas can be written as <span class=\"katex--inline\">\\frac{m}{n}</span>, where <span class=\"katex--inline\">m</span> and <span class=\"katex--inline\">n</span> are relatively prime positive integers. Compute <span class=\"katex--inline\">m+n</span>.</p>&#10;", "hints_md": "", "hints_html": "", "editorial_md": "", "editorial_html": "", "flag_hint": "", "point_value": 2, "problem_name": "Holiday Contest 2025 - Individual Round - Problem 2", "can_next": false, "can_prev": false, "nxt": "", "prev": ""}}