问题1.在一个班级中，有27个男孩和14个女孩。老师要选择1个男孩和1个女孩代表一个函数的班级。老师可以通过几种方式进行选择?
解决方案：
Given: Total number of boys = 27
Total number of girls = 14
So, ways to select a boy = 27 P1 = 27
Ways to select a girl = 14P1 = 14
Ways for selecting a pair of 1 boy, 1 girl = 27 x 14 = 378
问题2。一个人想从文具店购买一支钢笔，一支圆珠笔和一支铅笔。如果有10种钢笔品种，12种圆珠笔品种和5种铅笔品种，他可以从几种方式中选择这些物品?
解决方案：
Given: Total number of fountain pen = 10
Total number of ball pen = 12
Total number of fountain pencil = 5
Person wants to buy only one fountain pen, one ball pen, and one pencil
So, ways to select a pen = 10P1 = 10
Ways to select a ball pen = 12P1 = 12
Ways to select a pencil = 5P1 = 5
Ways for selecting the desired triplet = 10 x 12 x 5 = 600
问题3.从果阿到孟买，有两条路线；空气和海洋。从孟买到德里，共有3条路线；航空，铁路和公路。从果阿经孟买到德里，有多少种路线?
解决方案：
Given: From Goa to Bombay two routes = air, and sea
From Bombay to Delhi there are three routes = air, rail, and road
So, the routes from Goa to Bombay = 2P1 = 2
Routes from Bombay to Delhi = 3P1 = 3
Total different routes from Goa to Delhi = 2 x 3 = 6
问题4.薄荷糖准备金属日历，以月度表的形式指定月，日和日(每月一盘)。它应准备为将来的所有可能性准备多少种日历?
解决方案：
We need to find the different total number of calendars so that all the years can be represented by any one of these.
Case 1: Leap year
A leap year may start with any of 7 possible days (Monday to Sunday) = 7 options
Case 2: Ordinary year
An ordinary year may start with any of 7 possible days (Monday to Sunday) = 7 options
Total calendars = 7 + 7 = 14
问题5.有四个包裹和五个邮局。包裹可以通过几种不同的方式以挂号邮寄的方式寄出?
解决方案：
Given: Total number of parcels = 4
total number of post-offices = 5
Each of the four parcels have 5 options of post-offices.
So, each parcel can be sent in 5P1 ways.
Hence, the total ways = 5P1 x 5P1 x 5P1 x 5P1 = 5 x 5 x 5 x 5 = 625
问题6.投掷硬币五次，并记录结果。有多少可能的结果?
解决方案：
Each toss can result in 2P1 = 2 ways.
Five tosses = 2 x 2 x 2 x 2 x 2 = 32 ways of outcomes
问题7.考生可以通过几种方式回答一组十个正确/错误类型的问题?
解决方案：
For answering one question: 2 P1 = 2 ways
For answering 10 questions: 2 x 2 x 2 x……2 (10 times) = 210 = 1024 possibilities
问题8.字母锁由三个环组成，每个环上都标有10个不同的字母。有多少种方法可以尝试不成功地打开锁?
解决方案：
Number of possibilities for a single ring = 10 ways
For 3 rings: 10 x 10 x 10 = 1000 ways
Out of these, 1 way will be the correct password
So, the number of unsuccessful attempts = 1000 – 1 = 999
问题9.考试中有6个多项选择题。如果前三个问题每个有4个选择，接下来的三个问题每个有2个选择，那么答案的顺序可能是多少?
解决方案：
Possible sequences for first 3 questions = 4P1 x 4P1 x 4P1 = 4 x 4 x 4 = 64
Possible sequences for next 3 questions = 2P1 x 2P1 x 2P1 = 2 x 2 x 2 = 8
Total possibilities = 64 x 8 = 512
问题10.书店里有5本关于数学的书和6本关于物理学的书。一个学生可以用几种方式购买：
(i)一本数学书和一本物理书?
(ii)是一本数学书还是一本物理书?
解决方案：
Given: Total number of Mathematics book = 5
Total number of Physics book = 6
(i) Number of ways of buying Mathematics book = 5P1 = 5
Number of ways of buying Physics book = 6P1 = 6
Total possibilities = 5 x 6 = 30
(ii) Number of ways of buying a book (can be any Maths or Physics) = 11 P1 = 11
问题11.给定7个不同颜色的标志，如果一个信号需要使用两个标志，一个标志在另一个标志之下，则可以生成多少个不同的信号?
解决方案：
Ways to select 2 flags out of 7 = 7P2 = 7x(7 – 1)/2 = 21
Ways of generating different signals from these 2 selected flags = 2
(say, A and B are the colors selected then A can be above and B below; and vice versa so 2 ways)
Total distinct signals possible = 21 x 2 = 42
问题12：一个团队由6个男孩和4个女孩组成，另外一个团队有5个男孩和3个女孩。当一个男孩与一个男孩对战而一个女孩与一个女孩对战时，两队之间可以安排多少场单场比赛?
解决方案：
Case 1: A boy plays against a boy
Select a boy from team 1 and a boy from team 2
Team 1: 6P1 = 6
Team 2: 5P1 = 5
Total ways of a boy playing against a boy = 6 x 5 = 30
Case 2: A girl plays against a girl
Select a girl from team 1 and a boy from team 2
Team 1: 4P1 = 4
Team 2: 3P1 = 3
Total ways of a girl playing against a girl = 4 x 3 = 12
Total ways of signal matches = Ways of boy playing against boy + Ways of girl playing against girl
= 30 + 12
= 42
问题13。12名学生参加比赛。前三个奖项有多少种授予方式?
解决方案：
Number of ways of selecting 3 winners = 12P3 = 12 x 11 x 10 / (3 x 2 x 1) = 220
For 3 winners selected, different ways of assigning the position
For first position we have 3 possibilities of people,
then for second we have 2 possibilities (other than the one already given first position)
and for third we have 1 possibility (other than the ones declared second and first positions)
So, 3 x 2 x 1 = 6 possibilities of assigning these 3 positions to the three selected people
Total ways of giving 3 prizes = No. of ways of selecting 3 people x Assigning 3 positions to the 3 people
= 220 x 6
= 1320
问题14.有多少个具有10个项的AP，其第一个项在集合{1、2、3}中，并且其共同差值在集合{1、2、3、4、5}中?
解决方案：
Number of ways of selecting first term = 3P1 = 3
Number of ways of selecting common difference = 5P1 = 5
Total different A.P. series = 3 x 5 = 15
问题15.在大学的36名教师中，将任命一名校长，一名副校长和负责老师。有多少种方法可以做到这一点?
解决方案：
Number of ways of selecting 3 people = 36P3
= 36 x 35 x 34 / (3 x 2 x 1)
= 7140
For 3 people selected, different ways of assigning the posts
For the principal post we have 3 possibilities of people,
then for vice-principal post we have 2 possibilities (other than the one already given principal post)
and for teacher-in-charge we have 1 possibility (other than people declared principal and vice-principal)
So, 3 x 2 x 1 = 6 possibilities of assigning these 3 posts to the three selected people
Total ways of assigning 3 posts = No. of ways of selecting 3 people x Assigning 3 posts to 3 people
= 7140 x 6
= 42840
问题16.有多少个三位数的数字，没有重复的数字?
解决方案：
Ways of selecting 100th place = 9P1 = 9 (Selecting from all digits except 0)
Ways of selecting 10th place = 9P1 = 9 (Selecting from all digits except the digit placed at 100th position)
Ways of selecting unit pace = 8P1 = 8 (Selecting from all digits except those at 100th and 10th places)
Total 3-digit numbers possible with no digit repeated = 9 x 9 x 8 = 648