3, 這種題目叫分組問題,有正式公式去做,out of syllabus的,但用syllabus內的方法一樣做到。設12人為A, B, C, D, E, F, G, H, I, J, K, L,這題的做法是
1. 計出有序分組(ordered partition)的總數
2. 考慮有序分組數內有冇覆重成份,有要先修正,以計出正確的分組數目
3. 取一個分組case派發games,看看有多少種分派方法
4. 計答案
a) 首先計出有序分組的總數。方法是當不重置抽樣做
﹒ ﹒ ﹒ No. of ordered partitions = C(12, 3)C(9, 3)C(6, 3)C(3, 3)
跟往考慮分組數內有無重覆,方法是調轉兩組的次序諗。因抽得{ { A, B, C }, { D, E, F }, { G, H, I }, { J, K, L } }和{ { D, E, F }, { A, B, C }, { G, H, I }, { J, K, L } }當同一分組方法論,所以有重覆分組的情況發生,咁要除返重覆的次數以計出正確的分組數目
﹒ ﹒ ﹒ No. of diff partitions = No. of ordered partitions/4! = C(12, 3)C(9, 3)C(6, 3)C(3, 3)/4!
最後取一個特定分組如{ { D, E, F }, { A, B, C }, { G, H, I }, { J, K, L } }派games{ G1, G2, G3, G4 },計得
﹒ ﹒ ﹒ No. of ways in arranging games for a particular partition = 4!
推導出
﹒ ﹒ ﹒ No. of ways in grouping students to play the games
﹒ ﹒ ﹒ = No. of diff partitions×No. of ways in arranging games for a particular partition
﹒ ﹒ ﹒ = C(12, 3)C(9, 3)C(6, 3)C(3, 3)/4!×4!
﹒ ﹒ ﹒ = 369600 §
4. 設8人為A, B, C, D, E, F, G, H,如舊用上方法,先計出有序分組的總數
﹒ ﹒ ﹒ No. of ordered partitions = C(8, 2)C(6, 2)C(4, 2)C(2, 2)
跟往考慮分組數內有無重覆,方法是調轉兩組的次序諗。因抽得{ { A, B }, { C, D }, { E, F }, { G, H } }和{ { C, D }, { A, B }, { E, F }, { G, H } }當同一分組方法論,所以有重覆分組的情況發生,咁要除返重覆的次數以計出正確的分組數目
﹒ ﹒ ﹒ No. of diff partitions = No. of ordered partitions/4! = C(8, 2)C(6, 2)C(4, 2)C(2, 2)/4!
由於不用再編games, 推得
﹒ ﹒ ﹒ No. of ways in grouping students = No. of diff partitions
﹒ ﹒ ﹒ C(8, 2)C(6, 2)C(4, 2)C(2, 2)/4!
﹒ ﹒ ﹒ = 105 §
分組問題詳可睇
http://lsforum.net/board/thread-263646-1-9.html 我的解釋,咁會有幫助。
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本帖最後由 peterkcc2015 於 2016-3-12 06:20 PM 編輯 ]