组合和排列练习题|设置 1

先决条件：排列组合

n 学生出现在考试中，找出可以公布考试结果的方式。

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’ 场比赛将在国际象棋比赛中进行，找出可以决定结果的方法的数量

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

羽毛球锦标赛由“n”场比赛组成。
(i) 找出可以给出预测结果的方式的数量。
(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

找出“n”个不同的珠子可以排列成一条项链的方式数

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”个问题试卷，找出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

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排列组合测验
组合和排列练习题