送交者: jxh 于 2006-3-04, 12:20:56:

４个人分１５枚一模一样的硬币，如果每人至少要得一枚，一共有多少种分法？请列出你的算式。

• 楼下吴礼已经作出了精彩解答。鼓掌。 - jxh (0 bytes) 2006-3-05, 00:09:56
• 12! ? - SmilingKaKa~ (0 bytes) 2006-3-04, 23:49:21
  • NO. sigma(j=1)(12)sigma(i=1)(j) i = 364? - SmilingKaKa~ (0 bytes) 2006-3-05, 00:31:53
    • Sorry. Haven't checked WuLi's answer carefully. - SmilingKaKa~ (0 bytes) 2006-3-05, 00:35:53
      • Yeah, You guys are really smart. - SmilingKaKa~ (0 bytes) 2006-3-05, 00:53:41
• 4^15-4*3^15 - whoami (0 bytes) 2006-3-04, 20:06:18
  • gee, missed 一模一样 - whoami (0 bytes) 2006-3-04, 20:43:13
• c(14)(4) - zhangqq (7 bytes) 2006-3-04, 19:09:20
• 首先一人拿一枚 - 松鼠 (37 bytes) 2006-3-04, 12:24:23
  • 想法是好的 - testtest (61 bytes) 2006-3-05, 00:35:47
  • C(12)(3)=220种? - 匆匆过客 (0 bytes) 2006-3-04, 14:50:30
    • 12^3/3! = 288种? - 匆匆过客 (0 bytes) 2006-3-04, 15:19:47
  • 如何把11枚分给四个人？ - 吴礼 (132 bytes) 2006-3-04, 14:44:15
    • 高 - deadmeat (157 bytes) 2006-3-04, 15:09:15
      • perfect! - 松鼠 (0 bytes) 2006-3-05, 16:42:43
      • 好，简明。吴礼死肉厉害。 - jxh (0 bytes) 2006-3-04, 15:23:39
    • This is really smart way! - mangolasi (0 bytes) 2006-3-04, 15:01:44
      • I'll never think of these kind of tricks. - mangolasi (0 bytes) 2006-3-04, 15:03:11
        • I can't take too much credit - 吴礼 (78 bytes) 2006-3-04, 15:14:47
          • No, I've never seen this before. - mangolasi (49 bytes) 2006-3-04, 15:24:13
  • 不对。 - jxh (50 bytes) 2006-3-04, 12:42:03
    • Squirrel's first step is a good start. - mangolasi (226 bytes) 2006-3-04, 13:06:44
      • 还有个问题，你没考虑到4 groups里面有两个或以上的group的硬币数目相同的情况 - deadmeat (70 bytes) 2006-3-04, 13:39:10
        • Indeed. - mangolasi (209 bytes) 2006-3-04, 13:45:25
      • hmm, not a general formula yet, but for the first step: - mangolasi (545 bytes) 2006-3-04, 13:22:15
        • Forget to add a condtion of stopping taking: when the # you take - mangolasi (111 bytes) 2006-3-04, 13:28:05

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