Principle of tolerance and repulsion

The principle of tolerance and repulsion:

when counting, one must pay attention to no repetition, no omission, in order to make the overlapping part not to be repeated calculation, people have developed a new counting method. The basic idea is that the number of all objects contained in a certain content is calculated, and then the number of repeated calculations is excluded, so that the result of the calculation is neither omitted nor duplicated, and this counting method is called the principle of repulsion.

formula:

Example:

1. A school six ⑴ class has 45 students, each in the summer vacation to participate in the Sports training team, including the football team has 25 people, to participate in the volleyball team 22 people, to participate in the swimming team 24 people, football, volleyball have participated in 12 people, Football, swimming have participated in 9 people, volleyball, swimming, there are 8 people, Q: Three are the number of people to participate?

Analysis: The number of participants in the football team of 25 is a class of elements, the number of volleyball team 22 for the Class B elements, the number of participants in the swimming team of 24 is a class C element, both a class and Class B for the soccer volleyball 12 people, both Class B and C for the football swimming are 9 people to participate in, Both the Class C and Class A for the volleyball swimming 8 people, three are involved in a Class B class C sum set to X. Note: The title of each person to participate in the sports training team, so the total number of this class is a class B and Class C sum.

Answer: 25+22+24-12-9-8+x=45 solution to X=3

2. How many of the natural numbers from 1 to 1000 can be divisible by 3 or 5 ? How many of the numbers cannot be divisible by 3 or 5?

Analysis: Obviously, this is a Repeat Count problem (of course, if you are not afraid of trouble you can go to the number 3 multiples, 5 multiples). We can think of the "number divisible by 3 or 5" as the Class A and Class B elements, which can be " evenly divisible by 3 or 5 (multiples of 15)" is repeated The number of calculations, which is "both a class and a class B element". The number of elements of Class A or Class B is asked. We can not calculate directly, we must first find out the required conditions. 1000÷3=333 ... 1, the number that can be divisible by 3 is 333 (think about it, why?) In the same vein, you can find other conditions.

Principle of tolerance and repulsion