A group of seven people queue up to enter into a museum. It is given that Cherry and Rainbow are two of them.
If Cherry cannot be the first one and Rainbow cannot be the last one, how many arrangements are there?
Permutation and Combination (Advanced)?
2016-01-07 6:25 pm
回答 (3)
2016-01-09 2:49 am
✔ 最佳答案
the answer is wrong7! = 5040
5040 - 3720 = 1320
2016-01-08 1:46 pm
But the answer is 1320
2016-01-08 2:26 am
There are two complementary cases:
Case 1: Rainbow was the first.
Then there are no more restrictions for the remaining six persons, so the number of arrangements is 6!=720
Case 2: Rainbow was not chosen as the first
There are (7-Rainbow-Cherry)=5 choices for the first in line.
Next we choose the last person to enter (while lining up) with (6-Rainbow)=5 choices.
The remaining middle 5 positions can be filled in 5! ways for a total of 5*5*5! ways = 3000 ways.
Total number of ways = 720+3000 = 3720
Case 1: Rainbow was the first.
Then there are no more restrictions for the remaining six persons, so the number of arrangements is 6!=720
Case 2: Rainbow was not chosen as the first
There are (7-Rainbow-Cherry)=5 choices for the first in line.
Next we choose the last person to enter (while lining up) with (6-Rainbow)=5 choices.
The remaining middle 5 positions can be filled in 5! ways for a total of 5*5*5! ways = 3000 ways.
Total number of ways = 720+3000 = 3720
收錄日期: 2021-04-18 14:17:43
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https://hk.answers.yahoo.com/question/index?qid=20160107102517AANsCf5