什么是计数原理问题？

概率是数学的一个分支，用于计算事件发生的可能性的数值描述，或者我们可以说概率处理随机事件的发生。事件发生的概率总是在零和一之间。它可以用数字或百分比表示。在概率中，要找到特定事件的可能性，我们必须首先找到潜在结果的总数。

计数原理问题

计数原理问题基于简单的基本计数方法，牢记可用选项为特定选择选择合适的项目。计数原理问题最好使用基于树的结构方法来解决。可用选项表示为树的分支。树表示还简化了对计数原理问题的理解。计数问题可以最好地描述为：

If there are ‘n’ entities and each of the n entities has m1, m2, m3………………mn options to choose from. Say 1st entity has m1 choices, 2nd entity has m2 choices, 3rd entity has m3 choices and so on.

Then the total ways of the selection of entities would be :

Number of ways for Counting Principle Problems:

m1 x m2 x m3 x m4………………………………..x mn

学习计数原理问题的最好方法是通过一个例子：

Example: Consider Vaibhav has 3 mangoes, 3 papaya and 3 apples. In how many ways can he put fruit of one kind in a fruit basket.

Solution:

Then pairing can take place as follows:

(M1 P1 A1), (M1 P1 A2), (M1 P1 A3), (M1 P2 A1), (M1 P2 A2), (M1 P2 A3), (M1 P3 A1), (M1 P3 A2), (M1 P3 A3)

(M2 P1 A1), (M2 P1 A2), (M2 P1 A3), (M2 P2 A1), (M2 P2 A2), (M2 P2 A3), (M3 P3 A1), (M3 P3 A2), (M3 P3 A3)

(M3 P1 A1), (M3 P1 A2), (M3 P1 A3), (M3 P2 A1), (M3 P2 A2), (M3 P2 A3), (M3 P3 A1), (M3 P3 A2), (M3 P3 A3)

The total number of ways of choosing this pairing using Counting Principle Problems

Choices available for mangoes (m) = 3

Choices available for papaya (n) = 3

Choices available for apples (n) = 3

Total no. of ways: 3 X 3 X 3 = 27

类似问题

问题 1. 考虑 3 个男孩和 3 个女孩想要组队参加萨尔萨舞比赛。

解决方案：

Then pairing can take place as follows:

(B1 G1)    (B1 G2)  

(B2 G1)    (B2 G2)  

The total number of ways of choosing this pairing using Counting Principle Problems

Choices available for boys (m) = 2

Choices available for girls (n) = 2

Total no. of ways: 2 x 2 = 4

问题 2。考虑一个男孩有三种衬衫可供选择参加聚会。

解决方案：

The total number of ways of choosing this pairing using Counting Principle Problems:

There is one boy and three choices of shirt available.

Total no. of ways: 1 x 3 = 3

问题 3。假设有三个男孩，并且有三种衬衫可供选择参加聚会。

解决方案：

The total number of ways of choosing this pairing using Counting Principle Problems:

There are three boys and three choices of shirt available.

(B1 S1)  (B1 S2)  (B1 S3)

(B2 S1)  (B2 S2)  (B2 S3)

(B3 S1)  (B3 S2)  (B3 S3)

Total no. of ways: 3 x 3 = 9

问题 4. 考虑 Seema 从 4 种蔬菜选择和 2 种面包选择中进行选择。她有多少种方法可以结合蔬菜和面包来准备晚餐？

解决方案：

Choices available for vegetables: 4

Choices available for bread: 2

(V1 B1)  (V1 B2)

(V2 B1)  (V2 B2)

(V3 B1)  (V3 B2)

(V4 B1)  (V4 B2)

Total no. of ways: 4 x 2 = 8