组合和排列练习题|套装1

先决条件：排列组合

n名学生参加考试，找到可以宣布考试结果的方式。

Answer is 2ⁿ

Examples:
Input : n = 6
Output : Each student can either pass or fail in the examination. so ,there exists 2
possibilities for each of the 6 students in the result. hence total number of ways for the result=(2)⁶

Input : n = 8
Output :(2)⁸=256

“ n”类比赛将在国际象棋锦标赛上进行，找出决定其结果的方式

Answer is (3)ⁿ ways

Examples:
Input : n = 3
Output: The results of each of the 3 matches can be three ways namely win ,draw or loss
since total no. of ways in which results of 3 matches can be decided =(3)³

Input: 6
Output:(3)⁴=81

羽毛球比赛由’n’个比赛组成。
(i)找到给出可预测其结果的方式的数量。
(ii)包含所有正确结果的预测总数。
(ii)包含所有错误结果的预测总数。

Answer (i) (2ⁿ)
(ii) 1
(iii) 1
Examples:
Input : A badminton tournament consists of 3 matches.
(i) In how many ways can their results be forecast ?
(ii) How many different forecasts can contain all correct results ?
(iii) How many different forecasts can contain all correct results ?
Output:(i) Each badminton match can be decided in only 2 ways either win or
loss for a particular team so total number of ways results of 3
matches can be forecast=2³=8
(ii) Results of each match can be forecast wrong in only 1 way
Total no. of forecasts containing all wrong results = (1³) = 1
(iii) Similarly, result of each can be forecast correct in only 1 way.
total no .of forecasts containing all correct results = (1³) = 1

找到可以排列“ n”个不同珠子以形成项链的方式的数量

Answer is (n-1)!/2
Examples: For example 4 beads can be arranged in following ways.
….b1

b2…….b4

….b3

….b1

b3…….b2

….b4

….b1

b4…….b3

….b2

Since it does not matter where we place first bead. Total ways to arrange is (n – 1)!. But clockwise and anticlockwise arrangements are same, so total arrangements are (n – 1)!/2

有“ n”个问题论文，没有，学生可以尝试一个或多个问题的方式

Answer: (2ⁿ-1)ways.

For example a student will solve one or more questions out of 4 questions in following ways.
1) The student chooses to solve only one question, can choose in ⁴C₁
2) The student chooses to solve only two questions, can choose in ⁴C₂
3) The student chooses to solve only three questions, can choose in ⁴C₃
3) The student chooses to solve all four questions, can choose in ⁴C₄
So total ways are

⁴C₁ + ⁴C₂ + ⁴C₃ +⁴C₄
=2⁴-1 = 15 ways
We know sum of binomial coefficients from ⁿC₀ to ⁿC_n is 2ⁿ

有关排列和组合的更多实践问题：
排列组合测验
组合与排列练习题