Every station has tickets to other stations. When there are m stations, there will be m(m- 1) tickets. After adding n stations, when there are (m+n) stations, there will be (m+n) (m+n- 1) tickets. Then we can make the following statement:

(m+n)(m+n- 1)-m(m- 1)= 58

Simplify and get it

(m+n)(m+n- 1)-m(m- 1)= n(2m+n- 1)= 58

Since both m and n are integers, and 2m+n- 1 is also an integer, it can be concluded from the above formula that n and 2m+n- 1 are two factors of 58, and since 58=2*29= 1*58,

① n=2, 2m+n- 1=29, and m =14 can be obtained;

② n= 1，2m+n- 1=58，m = 29

So there are two answers: the original 14 station plus two new stations, or the original 29 stations plus a new station plus1; (This method is not called gathering or sprouting. This method is called analysis, full format, and end. )

The next two questions are about the priority arrangement of special elements or special posts.

Question 2:

Step by step, the first course is not a or b, and priority is given. Choose one from four, and five people in the other three courses will be arranged at random. There are a (5,3) = 60 choices, and there are always N=4*60=240 choices.

Question 3:

The last position (must be an even number) and the first position (must be non-zero) of a special position,

In the first category, the last digit is 0, and the other two digits are randomly selected with 4 digits, where n1= 4 * 3 =12;

In the second category, the last digit is 2 or 4, there are two cases, the first digit is not zero, there are three cases, and the middle position is randomly selected, there are three cases, so N2=2*3*3= 18 cases;

So there is always N=N 1+N2=30;

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