probability 急!!!

1)
Among the 10 books, put the book with the smallest serial number on the leftend. (1P1)
Among the 10 books, put the book with the largest serial number on the rightend. (1P1)
From the rest 8 books, arrange 7 books between the two ends. (8P7)

Number of arrangements
= 1P1 × 1P1 × 8P7
= 1 × 1 × (8!/1!)
= 40320


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2)
Select 9 books from the 10 books. (10C9)
Among the 9 books, put the book with the smallest serial number on the leftend. (1P1)
Among the 9 books, put the book with the largest serial number on the rightend. (1P1)
Arrange the rest 7 books between the two ends. (7P7)

Number of arrangements
= 10C9­ × 1P1 × 1P1× 7P7
= (10!/9!1!) × 1 × 1 × 7!
= 10 × 7!
= 50400

1) Required number of arrangements = 8C7 x 7! = 40320
因為10本書入面最細no.同最大no.必揀同fix死左一個係最左一個係最右,所以得番8本書可以再揀7本塞係中間,所以8C7 (8本揀7本)
7本排係中間既書可以rearrange,所以再乘7! (7本排序)


2) Required number of arrangements = 10C9 x 7! = 50400
呢題同1)既分別係10本揀左既9本入面,最細既fix死係最左邊,最大既fix死係最右邊。
例如:10 books -> 2 3 4 5 6 7 8 9 10 left end:2 right end:10
10 books-> 1 2 3 4 5 6 7 8 9 left end:1 right end:9
所以10C9 (10本揀9本)
然後中間個7本就可以rearrange,所以再乘7! (7本排序)