✔ 最佳答案
The number of ways to arrange A, B, C, D, E, F & G randomly = 7! = 5040
(a) A before C
It is equal chance for A before or after C,
therefore, the no. of arrangement that A before C = 5040 / 2 = 2520
(b) A before C, and C before E
Just consider A, C & E, there are 6 different sequence of arrangement.
therefore, the cases for A before C, and C before E = 5040 / 6 = 840
(c) All cases in (b) should have C after A.
So, I assumed the question should be: C immediate after A
If so, we may group AC together, and consider to be 1 new letter.
Then, now we have only 6 letters : AC, B, D, E, F & G
The arrangement for these 6 letters = 6! = 720
However, these 720 arrangement contain AC before E, and AC after E.
Therefore, the number for arrangement with AC before E = 720/2 = 360
and all these 360 cases are applied to (b).
2006-12-04 23:44:08 補充：
補充 part (b) 點解 除6:由於 part (b) 的要求只 要 A在C之前, 各C在E之前,所以其他的字母並不構成影響,可以不作考慮。只考慮 A,C,E 3個字母, 其排序有6個不同組合的先後次序,而我們只抽出順序為 A,C,E 的1項.所以可能性是所以組合的 6分之1.
