✔ 最佳答案
Case 1 : Both parents sit at theends
No. of ways that the parents sit at the ends = P(2,2)
No. of ways that the 5 children sit between the ends = P(5,5)
Hence, no. of ways for case 1 = P(2,2) x P(5,5)
Case 2 : Only one parent sit at one end
No. of ways to select 1 parent from Tom and Mary = C(2,1)
No. of ways that this parent sit at one of the ends = P(2,1)
No. of ways that another parent sit between the ends = P(5,1)
No. of ways that the 5 children take their seat = P(5,5)
Hence, no. of ways for case 1 = C(2,1) x P(2,1) x P(5,1) x P(5,5)
Total number of ways
= P(2,2) x P(5,5) + C(2,1) x P(2,1) x P(5,1) x P(5,5)
= 2! x 5! + (2!/1!1!) x (2!/1!) x (5!/4!) x 5!
= 2 x 120 + 2 x 2 x 5 x 120
= 2640
2012-04-30 05:23:21 補充:
Case 1 並沒有漏計,而是你對英文字 "ends" 的理解錯誤。
"ends" 是眾數,意思是「兩端」。
即 Mary x x x x x Tom
或 Tom x x x x x Mary
所以 Mary 和 Tom 坐法有 P(2,2) 種,不用乘以 2。
如你提的坐法: Mary Tom x x x x x
Mary 坐在 one end,而 another end坐的是 x,不合題意。
同理,另一坐法: x x x x x Mary Tom
坐在 two ends 的是 x 和 Tom,不合題意。
而樓下的梁小姐可能亦對英文字 "ends" 理解錯誤,所以計算錯了,多乘了 2。
2012-04-30 05:27:41 補充:
這課題的題目通常有幾種解法,各適其式吧。我的答案因為附有解釋,所以較長。但真正回答時,只用寫:
Total number of ways
= P(2,2) x P(5,5) + C(2,1) x P(2,1) x P(5,1) x P(5,5)
= 2640