Output 1234 three digits without repetition

introduction

When solving mathematical permutations , Ask to give when encountering 1,2,3,4 When four numbers form repeated three digit numbers , Similar problems can be solved by programming , It can also verify whether the answer is correct .

1 problem

use 1,2,3,4 Four numbers form three digits without repetition , Find how many such numbers can be formed .

2 Method

Give a list first , The first cycle gets the first number , The second cycle gets the second number , The third cycle gets the third number , use if Conditional statements are used to judge whether the three numbers are repeated or equal , Then convert it to three digits , Add to list , Finally, output the list and get the result .

3 Experimental results and discussion

Through the experiment 、 Practice has proved that the proposed method is effective , Is able to solve the problem raised at the beginning .

Code list 1

List=[] for i in range(1,5):  for m in range(1,5): for i in range()1,5:  if i!=m and i!=n and m!=n:    x=(i*100+m*10+n)    list.append() print(list) print(‘ Can generate %d individual ’%int(len(list)))

4 Conclusion

Discussed 1234 How many non repeating three digits can be formed , involves for loop 、if Conditional judgment statements and conversion between strings , These knowledge points have been further consolidated , The result of permutation and combination by mathematical method is consistent with the result of the program , It is proved that this method is effective to solve this problem .