{"height":"190","published":"2008-04-27 11:03:49","description":"40\u554f\u9054\u6210\uff01\uff01Problem 53 - Project Euler There are exactly ten ways of selecting three from five, 12345:123, 124, 125, 134, 135, 145, 234, 235, 245, and 345In combinatorics, we use the notation, 5C3 = 10.In general,nCr = n!/r!(n\u2212r)!,where r \u2264 n, n! = n\u00d7(n\u22121)\u00d7...\u00d73\u00d72\u00d71, and 0! = 1.It is not until n = 23, th\u2026","author_url":"https://blog.hatena.ne.jp/tanakaBox/","provider_name":"Hatena Blog","html":"<iframe src=\"https://hatenablog-parts.com/embed?url=https%3A%2F%2Fboxnos.hatenablog.com%2Fentry%2F20080427%2F1209261829\" title=\" Problem 53 - \u7d44\u307f\u5408\u308f\u305b - \u30dc\u30af\u30ce\u30b9\" class=\"embed-card embed-blogcard\" scrolling=\"no\" frameborder=\"0\" style=\"display: block; width: 100%; height: 190px; max-width: 500px; margin: 10px 0px;\"></iframe>","categories":["Scheme"],"blog_url":"https://boxnos.hatenablog.com/","url":"https://boxnos.hatenablog.com/entry/20080427/1209261829","title":" Problem 53 - \u7d44\u307f\u5408\u308f\u305b","blog_title":"\u30dc\u30af\u30ce\u30b9","image_url":null,"provider_url":"https://hatena.blog","type":"rich","version":"1.0","author_name":"tanakaBox","width":"100%"}