permutation and combination

[匿名]

2010-11-19 11:23 pm

1.) Amy,Bobo and their 5 other friends are arranged to stand a row. Find the
no of arrangments if Amy can't be the first one and Bobo can't be the last
one.

2.) There are 12 different balls and 5 of them are blue color.The balls are now
arranged in a row.

How many permulations are there :

i. if two of the blue balls can't be placed next to each other ?
ii. if only 3 of the blue balls are placed next to each other ?
iii. if the blue balls must be placed at the two ends ?
iv. if the blue balls can't be placed at the two ends ?
v. if the blue balls can't be placed at both ends at the same time ?

3.) There are 9 different books.

How many ways to group them :

i) group into 3 piles ,such that 2 books in a pile,3 books in another piles,
4 books in remaining pile.

ii) groups into 3 piles,such that there are 3 books in each pile.

How many ways to distribute the books

iii) to distribute the books to Alan,Bobo,Carmen,such that one of them
gets 2 books,one gets 3 books,and the remaining one gets 4 books.

iv) to distribute the books to David,Eva and Fanny,such that each of them
gets 3 books.

v.)How many ways to arrange the books in a shelf with 3 racks,such that
there are 3 books on each racks?

Pls explain details.
THX A LOT!

回答 (1)

用戶9112

2010-11-20 6:06 am

✔ 最佳答案

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圖片參考：http://img23.imageshack.us/img23/5072/44598336.png

2010-11-19 22:06:38 補充：
http://img23.imageshack.us/img23/5072/44598336.png

2010-11-19 23:46:19 補充：
Amendment
(1)Amy+Bobo+5 other friends=total 7 persons
Arrangement=7!-6!-6!+5!=3720
(2)(i)No 2 blue balls together=C(8,5)*7!*5!=33868800
(ii)Find the 3 slots for the 3 blue ball groups=C(8,3)=56
Arrangement of the blue balls group in these position=3!=6
To choose 3 balls from among 5= C(5,3)=10

2010-11-19 23:46:36 補充：
Arrangement of the 3 blue balls within the group=3!=6
Arrangements of the non-blue balls=7!=5040
Total number of ways=56*6*10*6*5040=101606400
(iii) Probably the question means 1 blue balls at each end
There are C(5,2)*2=20 ways
Consider permutation of the rest 10 balls becomes 20*10!= 72576000

2010-11-19 23:46:52 補充：
(iv) Each end has 1 non-blue balls
Number of ways=C(7,2)*2*10!= 152409600
(v) Can have blue balls at both end=12!-72576000= 406425600

2010-11-19 23:47:01 補充：
(3)(i) since the arrangement of the books with the piles are not important,
So number of ways=362880/2!/3!/4!=1260
(ii) arrangement within pile not important so 60480/3!/3!/3!=280
(iii) correct # of ways=1260*6=7560
(iv) correct # of ways=280*6=1680

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