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课程表排列

发布时间: 2020-11-29 06:58:36

1. 课表排列,语,数,物,化,英,体六节课,数,必须排在体前边,则共有多少种排法

六节课任意排是A(6,6),由于数体是等可能排在前面的,因此有A(6,6)/2种排法,即6*5*4*3*2*1/2=360

2. 为什么课程表上的语文、数学要分开排列的道理

什么叫分开排列。。。
难道你要一堂课语文数学2老师一起教。。。

3. 请问课程表怎么排列,有捷径吗

网上有排课软件

4. 我想打印课表,在word中,课表是竖着排列的,我想换个角度,成横排列,怎样操作

可以通过菜单--文件--页面设置--页边距--方向,选择横向来等效实现。

5. 排课表——排列组合的问题

不考虑题目限制,总方法数为P(8,8)=40320种课表;

其中限定的某节课为某科目的情况(如第一节为英语)有P(7,7)=5040种课表,这样的限制条件的组合有C(4,1)=4种,所以共有5040*4=20160种课表;

其中限定的某两节课为某两个科目的情况(如第一节为英语、第二节为语文)有P(6,6)=720种课表,这样的限制条件的组合有C(4,2)=6种,所以共有720*6=4320种课表;

其中限定的某三节课为某三个科目的情况(如第一节为英语、第二节为语文、第三节为物理)有P(5,5)=120种课表,这样的限制条件的组合有C(4,3)=4种,所以共有120*4=480种课表;

其中限定的某四节课为某四个科目的情况(如第一节为英语、第二节为语文、第三节为物理、第四节为体育)有P(4,4)=24种课表,这样的限制条件的组合有C(4,4)=1种,所以共有24*1=24种课表;

所以最终答案为:40320-20160+4320-480+24=24024种

反复加减是因为在前一步中有重复计算的情况。

6. 排列组合关于课表问题

(1):第一节不排体育,则第一节的排法有5种:即C5(下面)1(上面),也可以写成内A5(下面)1(上面);最后一节不排数学,容则最后一节有4种排法,因为排第一节时用了一颗:C4(下面)1(上面);最后剩下中间的4节课,则是A4(下面)4(上面);总的排法就为:C51*C41*A44
(2)假设第一节是物理:则第二节的排法为:A51或C51;第六节的排法为:A41或C41;中间的为:A33;这种情况的排法为:A51*A41*A33
当第一节不是物理时:则第一节的排法为A41或C41;第二节的排法为:A41或C41;第六节的排法为:A31或C3;中间的为:A33;这种情况的排法为:A41*A41*A33
全部的排法为A41*A41*A33+A51*A41*A33

7. asp课程表排列代码

定义一个主键然后关联起来就可以了啊

8. SQL语句的一道题 三个基本表:学生表(Student)、课程表(Course)、学生选课表(SC)

1. select * from SC
2. select Sname,Sage from Student where Sdept = '计算机'
3. select Sno,Cno,Grade from SC where Grade >= 70 and Grade <= 80
4. select Sname,Sage from Student where Sage between 18 and 20 and Ssex = '男'
5. select top 1 Grade from SC where Cno = 'C01'
6. select max(Sage),min(Sage) from Student
7. select Sdept,sum(Sno) from Student group by Sdept
8. select course.Cname,sum(sc.Sno),max(Grade) from SC
join studet on Student.Sno = SC.Sno
join Course on Course.Cno = SC.Cno
group by course.cname,max(grade)

9. select sum(Cno),avg(Grade) from SC
join Course on Course.Cno = SC.Cno
join Student on Student.Sno= SC.Sno
order by SC.Sno

10. select Stuent.Sno,Stuent.Sname,sum(Grade) A from SC
join Student on Student.Sno = SC.Sno
group by sc.Sno,student.Sname
having A > 200

11. select Student.Sname,Student.Sdept from Student
join Course on Course.Cno = SC.Cno
join SC on SC.Sno = Student.Sno
where SC.Cno = 'C02'

12. select Student.sname,course.cno,sc.grade from sc
join student on student.sno = sc.sno
join course on course.cno = sc.cno
where sc.grade >= 80
order by sc.grade desc

13. select cno,cname from
(
select course.cno,course.cname,sun(sno) from student
join course on course.cno = sc.cno
join sc on sc.sno = student.sno
group by cno,cname
having sun(sno) > 0
)

14. ① select student.sname,student.sdept from
(
select student.sname,student.sdept,course.cname from student
join sc on sc.sno = student.sno
join course on course.cno = sc.cno
where course.cname = 'C01'
)

② select student.sno,student.sname from
(
select student.sno,student.sname,student.sdept,sc.grade from sc
join student on student.sno = sc,sno
where student.sdept = '信息' and sc.grade >= 80
)

③ select top 1 student.sname from
(
select student.sname,student.sdept,sum(sc.grade) from sc
join student on student.sno = sc.sno
where student.sdept = '计算机'
group by student.sname,student.sdept
order by
)

15. delete from sc where grade < 50
16. update sc set grade += 5 from sc
join course on course.cno = sc.cn
join student on student.sno = sc.sno
where student.sno =
(
select student.sno from sc
join course on course.cno = sc.cno
where course.cname = 'c01'
)

17. update sc set grade += 10 from sc
join student on student.sno = sc.sno
where student.sno =
(
select student.sno from student
join sc on sc.sno = student.sno
join course on course.cno = sc.cno
where student.sdept = '计算机' and course.cname = '计算机文化基础'
)

18. create view [A] as
select student.sno,student.sname,student.sdept,course.cno,course.cname,sc.grade from sc
join student on student.sno = sc.sno
join course on course.cno = sc.cno

19. create view [A] as
select student.sno,avg(sc.grade) from sc
join student on student.sno = sc.sno
group by student.sno

20. create view [A] as
select student.sno,sum(sc.grade) from sc
join student on student.sno = sc.sno
group by student.sno

21. create index A on student(sname)
22. 不会

9. 学校某天有6节不同的课,其中一节语文,一节数学必须排在前三节,这一天的课表有几种排法

这节课前三节如果都是不同的,就是6×5×四,如果可以相同的就是6×6×六,也就是六的立方。

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