permutation and combination20分

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

Required No. of ways = 7! - ( 5 x 4 ) x 5! = 5040 - 2400 = 2640

Here 7! means no. of ways they sit randomly

( 5 x 4 ) x 5! means no. of ways the parents both NOT sitting at the ends.

Is it the easiest method?

2012-04-30 07:43:06 補充：
At two ends means either Left or Right only.

If they both sit at two ends, that means one at Left and one at Right but they two cannot sit together as a pair on either one side. There is a misunderstanding.

wanszeto 啱。

但我用的方法更簡單。

首先，7人任坐有7！＝5040款

跟住，其中有一些是家長均沒有在兩側的：是5P2 x 5! = 2400款
5P2：中間位5個（扣除最左最右兩個），Tom & Mary 坐任意2個（其餘3個空出），有20款。
5!：5個細路坐落其餘剩餘座位。
所有可能減去父母均不在兩側，便是"最少一個在邊位上"了：
5040 - 2400 = 2640款

若覺得我幫到你，請可否到我的 Facebook Page 比個 Like 我：
經數學人
http://fb.com/ronaldchik