# Topic: There are four digits: 1, 2, 3 and 4.
# How many different three digits can be formed without repeating numbers? How much is each?


# Procedure analysis: traverse all possibilities and shave out duplicates.

total = 0
for i in range(1,5):
    for j in range(1,5):
        for k in
 
 range(1,5):
            if i!=j and j!=k and k!=i:
                print(i,j,k)
                total += 1
print(total)



# Simple Method: Use permutations in itertools

import itertools

sum2 = 0
a = [1,2,3,4]
for i in itertools.permutations(a,3):
    print(i)
    sum2 += 1
print(sum2)

#    permutations method emphasizes permutations

import itertools
n=int(raw_input())
a=[str(i) for i in range(n)]
s=""
s=s.join(a)
for i in itertools.permutations(s,n):
    print ‘‘.join(i)

#combinations method focuses on combination

import itertools
n=int(raw_input())
a=[str(i) for
 
 i in range(n)]
s=""
s=s.join(a)
for i in itertools.combinations(s,n):
    print ‘‘.join(i)

Python_每日習題——0001