There are k ways to accomplish one thing, including n 1, N2...NK method.
* * * There is N 1+N2+...+NK method.
2. Multiplication principle
To accomplish one thing, it takes k steps. There are m 1 and M2 ... the MK method of each step.
A * * * has m 1? m2？ ……? Mk method
arrange
Choose n objects from m different objects in turn and arrange them in a row. A * * * has a (n, m) species.
A(n，m)= m(m- 1)……(m-n+ 1)= m！ /(m-n)！
combine
Choose n (out of order) from m different objects to form a set, and one * * * has C(n, m).
C(n，m)= m(m- 1)……(m-n+ 1)/n！ =m！ /(m-n)！ n！
Example 1
How many different five-digit even numbers can 0, 1, 2, 3 and 4 form?
A(4，4)+3？ A(3，3)+3？ A(3，3)=60
Answer: 60 kinds.
Example 2
How many 6-digit numbers are composed of 2 1, 3 2 and 1 3?
C(2，6)？ C(3，4)？ C( 1， 1)=60
Answer: 60 kinds.
conventional process
(1) enumeration method
(2) Calculation with permutation number and combination number, including case discussion, application of inclusion and exclusion principle, and repeated consideration.
(3) Turn the problem into a general situation and then solve it with recursive thinking.
(4) Construct a combined model, and then solve it with corresponding ideas.
5. Recursive method
Example 3
10 How many areas can a plane be divided into? (Idea: Generalize the problem)
n= 1 2 3 4
m=2 4 7 1 1
a(n)=a(n- 1)+n
a(5)= 16 a(6)=22
a(7)=29 a(8)=37
a(9)=46 a( 10)=56
Answer: 56
6. Correspondence method (building geometric model)
For example: inserting board method, marking method (shortest route)
Example 4
How many positive integer solutions does the indefinite equation x1+x2+x3+x4+X5+X6 =10 have? (Using the plug-in method)
Take 10 balls and divide them into 6 groups.
OO | OO | O | OOO | O | O
x 1 x2？ x3 x4 x5 x6
That is, the problem is transformed into inserting five partitions in nine air spaces.
C(5，9)= 126
Conclusion 1
The positive integer solution of x 1+x2+ ... +xn = m has C(n- 1, m- 1) groups.
Conclusion 2
Several groups of nonnegative integer solutions of x 1+x2+ ... +xn = m?
→(x 1+ 1)+(x2+ 1)+……+(xn+ 1)= m+n
That is, the problem is transformed into a positive integer solution of y 1+y2+ ...+yn = m+n.
That is, C(n- 1, m+n- 1)
It's not easy to organize, thank you.