python possible combinations

1itertools.combinations(iterable, r)

1# 1. Print all combinations 
2from itertools import combinations
3
4comb = combinations([1, 1, 3], 2)
5print(list(combinations([1, 2, 3], 2)))
6# Output: [(1, 2), (1, 3), (2, 3)]
7
8# 2. Counting combinations
9from math import comb
10print(comb(10,3))
11#Output: 120
12

1all_combinations = [list(zip(each_permutation, list2)) for each_permutation in itertools.permutations(list1, len(list2))]

1import math
2n=7
3k=5
4print(math.comb(n, k))

1# A Python program to print all
2# permutations using library function
3from itertools import permutations
4 
5# Get all permutations of [1, 2, 3]
6perm = permutations([1, 2, 3])
7 
8# Print the obtained permutations
9for i in list(perm):
10    print (i)
11

1def permutations(iterable, r=None):
2    # permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC
3    # permutations(range(3)) --> 012 021 102 120 201 210
4    pool = tuple(iterable)
5    n = len(pool)
6    r = n if r is None else r
7    if r > n:
8        return
9    indices = list(range(n))
10    cycles = range(n, n-r, -1)
11    yield tuple(pool[i] for i in indices[:r])
12    while n:
13        for i in reversed(range(r)):
14            cycles[i] -= 1
15            if cycles[i] == 0:
16                indices[i:] = indices[i+1:] + indices[i:i+1]
17                cycles[i] = n - i
18            else:
19                j = cycles[i]
20                indices[i], indices[-j] = indices[-j], indices[i]
21                yield tuple(pool[i] for i in indices[:r])
22                break
23        else:
24            return
25