# 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)