举🌰：

五个盒子，每个盒子里面有3个球（数字表示），每次每个盒子取一个，求组合数，球编号之和。

方法一：for 循环--多层嵌套

box,box1,box2 =  [0,1,2],[0,3,4],[0,5,6]    # 0，1，2 三个盒子，数字代表球的种类
box3,box4,box5 = [0,5,6],[0,3,4],[0,1,2]    # 3，4，5 三个盒子
Box = [box,box1,box2,box3,box4,box5]
count = 0  
# methond 1
for a in Box[0]:
    for b in Box[1]:
        for c in Box[2]: 
            for d in Box[3]:
                for e in Box[4]:
                    for f in Box[5]:
                        print((a,b,c,d,e,f),sep = ',')
                        sum = a+b+c+d+e+f
                        print("sum:",sum)
                        count  += 1
print("count:",int(count))



# 运行结果(展示一部分）
..............
(2, 4, 6, 6, 3, 2)
sum: 23
count: 726
(2, 4, 6, 6, 4, 0)
sum: 22
count: 727
(2, 4, 6, 6, 4, 1)
sum: 23
count: 728
(2, 4, 6, 6, 4, 2)
sum: 24
count: 729

方法一代码太low了，脸上挂不住啊～

方法二：python内置itertools迭代器，具体介绍可以看一下这位小哥的博客：

https://blog.csdn.net/u013300049/article/details/79313979

import itertools

box,box1,box2 =  [0,1,2],[0,3,4],[0,5,6]
box3,box4,box5 = [0,5,6],[0,3,4],[0,1,2]
Box = [box,box1,box2,box3,box4,box5]
count = 0                  
# methond 2
for i in itertools.product(box,box1,box2,box3,box4 ,box5):
    print(i)
    count += 1
    print("sum:",sum(i))
    print("count:",count)


# 结果（展示部分）
..............
(2, 4, 6, 6, 3, 2)
sum: 23
count: 726
(2, 4, 6, 6, 4, 0)
sum: 22
count: 727
(2, 4, 6, 6, 4, 1)
sum: 23
count: 728
(2, 4, 6, 6, 4, 2)
sum: 24
count: 729

总结：如果这篇博客让你有多一点点了解编程知识，那就点个赞吧！