Use Python’s standard library to generate or count selections: itertools.permutations() is for ordered selections without replacement, and itertools.combinations() is for unordered selections without replacement. If items may repeat, use itertools.product() or itertools.combinations_with_replacement(), depending on whether order matters. If you only need the number of outcomes, use math.perm() or math.comb() instead of generating them.
Permutations vs. combinations: does order matter?
A permutation is an ordered selection. From A, B, and C, selecting two gives AB, AC, BA, BC, CA, and CB. AB and BA are different outcomes.
A combination is an unordered selection. Choosing two of the same three items gives AB, AC, and BC; reversing a pair does not create a new selection.
Think of race positions as a permutation because first and second place differ. Choosing members for a committee is a combination because the order of the chosen members does not matter. Order is only part of the decision, though: you must also decide whether an item can be selected more than once.
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Choose the right Python function
| What you need | Python tool | Order matters? | Can an input position repeat? |
|---|---|---|---|
Generate ordered selections of length r |
itertools.permutations() |
Yes | No |
Generate unordered selections of length r |
itertools.combinations() |
No | No |
| Generate unordered selections with repetition | itertools.combinations_with_replacement() |
No | Yes |
| Generate ordered sequences with repetition | itertools.product() |
Yes | Yes |
| Count ordered selections without replacement | math.perm() |
Yes | No |
| Count unordered selections without replacement | math.comb() |
No | No |
| Get one random selection without replacement | random.sample() |
Returned order is meaningful | No |
| Randomly rearrange an entire list | random.shuffle() |
Yes | No |
For example, a PIN where digits can repeat is an ordered sequence with repetition, so product() fits. If digits cannot repeat, use permutations. For topping quantities where order does not matter and a topping can be chosen more than once, use combinations with replacement.
Generate permutations with itertools.permutations()
Import permutations from itertools. Its optional r argument sets the selection length; if omitted, Python generates full-length permutations.
from itertools import permutations
items = ["A", "B", "C"]
for result in permutations(items, 2):
print(result)Output:
('A', 'B')
('A', 'C')
('B', 'A')
('B', 'C')
('C', 'A')
('C', 'B')Each result is a tuple. The function returns an iterator, so results can be handled one at a time rather than stored together. Its count for n input positions and selection length r is n! / (n - r)!. For three inputs and two positions, that is 3 × 2 = 6.
Generate combinations with itertools.combinations()
Use combinations(iterable, r) when selecting r items without replacement and the order of selection does not matter.
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from itertools import combinations
items = ["A", "B", "C"]
for result in combinations(items, 2):
print(result)Output:
('A', 'B')
('A', 'C')
('B', 'C')The pair ('B', 'A') is not emitted separately: it represents the same combination as ('A', 'B'). The count is n! / (r! × (n - r)!); for three inputs taken two at a time, it is three.
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Both functions accept iterables such as strings, lists, tuples, and ranges, and yield tuples. Their output follows the input order; they do not automatically sort values alphabetically. The official Python documentation for permutations() and documentation for combinations() describes these behaviors. The iterable is consumed into a tuple internally, so these functions are not appropriate for a genuinely unbounded input.
Count outcomes without generating them
When you need only a count, use the math module. math.perm(n, r) counts ordered selections without replacement, while math.comb(n, r) counts unordered selections without replacement.
from math import perm, comb
perm(10, 3) # 720
comb(10, 3) # 120These functions were added in Python 3.8. They accept integer arguments and raise ValueError for negative arguments. If r is greater than n, both return 0. The corresponding iterators also yield no tuples when asked to select more positions than the input contains. For r = 0, each iterator yields one empty tuple, (): there is exactly one way to select nothing.
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from math import factorial
def permutation_count(n, r):
return factorial(n) // factorial(n - r)
def combination_count(n, r):
return factorial(n) // (factorial(r) * factorial(n - r))Counting first is a useful feasibility check. For example, perm(10, 10) is 3,628,800, while comb(50, 6) is 15,890,700. Those counts indicate why generating and storing every result can become impractical.
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Allow repetition: products and combinations with replacement
Ordered sequences with repetition
Use product(iterable, repeat=r) when every position independently draws from the same pool and order matters. It is the Cartesian product of the pool with itself r times.
from itertools import product
list(product("AB", repeat=2))
# [('A', 'A'), ('A', 'B'), ('B', 'A'), ('B', 'B')]For n choices at each of r positions, there are n ** r sequences. Two positions and two symbols therefore produce 2 ** 2 = 4 outcomes. Unlike permutations, products allow a pool position to be used again.
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Use combinations_with_replacement(iterable, r) when order does not matter but a pool position may be selected more than once.
from itertools import combinations_with_replacement
list(combinations_with_replacement("ABC", 2))
# [('A', 'A'), ('A', 'B'), ('A', 'C'),
# ('B', 'B'), ('B', 'C'), ('C', 'C')]The number of outcomes for n input positions taken r at a time with replacement is comb(n + r - 1, r). For three choices taken two at a time, that is comb(4, 2) = 6. The official product() documentation and combinations_with_replacement() documentation provide further details.
Get a random selection instead of every possibility
If you need one random sample without replacement, random.sample() returns a list. Its returned order is a random ordered selection.
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import random
items = ["A", "B", "C", "D"]
ordered_sample = random.sample(items, k=3)If the selection is conceptually a combination, normalize its display order so a pair appears consistently rather than sometimes as ('A', 'B') and sometimes as ('B', 'A'):
unordered_sample = tuple(sorted(random.sample(items, k=2)))Use random.shuffle(items) to randomly reorder an entire list in place; it changes items. For a large integer population, a range can be sampled efficiently, for example random.sample(range(10_000_000), k=60). See the Python documentation for random.sample() and random.shuffle().
The standard random module uses the deterministic Mersenne Twister and is not suitable for cryptographic uses such as security-sensitive tokens or passwords. Use the secrets module for cryptographic randomness; gambling systems may also have specialized security and regulatory requirements beyond ordinary pseudorandom sampling.
Handle duplicate values deliberately
itertools treats input positions as distinct, even when their values are equal. Thus permutations("AAB", 2) generates six tuples, including two copies each of ('A', 'A'), ('A', 'B'), and ('B', 'A'). This is correct when the two A characters are distinct physical positions, but it can be surprising if the desired output is unique by value.
For a small result set, convert to a set to remove duplicate tuples after generation:
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from itertools import permutations
unique_results = set(permutations("AAB", 2))
# {('A', 'A'), ('A', 'B'), ('B', 'A')}This still generates the duplicates first. For larger inputs, a frequency-aware generator can avoid producing duplicate-by-value permutations:
from collections import Counter
def unique_permutations(values, r=None):
counts = Counter(values)
r = len(values) if r is None else r
def build(path):
if len(path) == r:
yield tuple(path)
return
for value in counts:
if counts[value] == 0:
continue
counts[value] -= 1
path.append(value)
yield from build(path)
path.pop()
counts[value] += 1
yield from build([])
list(unique_permutations("AAB", 2))
# [('A', 'A'), ('A', 'B'), ('B', 'A')]For this implementation, the values must be usable as dictionary keys because Counter stores them as keys.
Process large result spaces without wasting memory
Iterate instead of materializing everything
A loop can process each result as it is produced:
from itertools import permutations
for result in permutations(items, 3):
if result[0] == "A":
print(result)Avoid converting an iterator to a list unless you know the output is small enough to store. The iterator avoids holding all tuples at once, but it does not reduce the total work required if your program consumes every tuple. Full permutations grow as n!; length-r permutations grow as n! / (n - r)!; combinations grow as comb(n, r); products grow as n ** r; and combinations with replacement grow as comb(n + r - 1, r).
Take only a preview or first few results
Use itertools.islice() to stop after a fixed number of results without building the full list:
from itertools import islice, permutations
first_five = islice(permutations(range(10), 3), 5)
for result in first_five:
print(result)Filter results or prune earlier
Filtering with a generator expression keeps the filtered results lazy:
from itertools import permutations
items = ["A", "B", "C", "D"]
valid = (
result for result in permutations(items, 3)
if result[0] != "D"
)
for result in valid:
print(result)This still constructs and checks every candidate permutation before deciding whether it passes the condition. When constraints make many partial candidates impossible, backtracking can prune those branches early. For example, this generator produces ordered selections without replacement and lets a condition be added before the recursion continues:
def arrangements(items, r):
def build(path, remaining):
if len(path) == r:
yield tuple(path)
return
for index, item in enumerate(remaining):
next_path = path + [item]
next_remaining = remaining[:index] + remaining[index + 1:]
# Add a constraint check here before descending.
yield from build(next_path, next_remaining)
yield from build([], list(items))For ordinary unconstrained generation, prefer the concise built-in iterator. Custom backtracking is useful when it can reject an invalid partial path before exploring all of its possible extensions.
Quick Recap
Common mistakes to avoid
- Using permutations when order does not matter creates redundant outcomes; use combinations instead.
- Using combinations for a code or sequence where position matters omits valid orderings; check whether repetition is allowed and use permutations or product accordingly.
- Assuming permutations or combinations reuse an input position: neither does. Choose a replacement-enabled function if repetition is required.
- Calling
len(list(permutations(...)))just to count: usemath.perm()ormath.comb()for the relevant no-replacement count. - Assuming duplicate values are automatically collapsed: standard
itertoolsfunctions distinguish input positions. - Assuming combinations sort input values: results follow the input sequence; sort the input first only if that is the intended output order.
- Using
randomfor passwords, tokens, or other security-sensitive randomness: usesecrets.
Quick reference
| Goal | Example | Meaning |
|---|---|---|
| Ordered selection, no replacement | permutations("ABC", 2) |
Six ordered pairs |
| Unordered selection, no replacement | combinations("ABC", 2) |
Three pairs |
| Ordered sequence, with repetition | product("AB", repeat=2) |
Four sequences |
| Unordered selection, with repetition | combinations_with_replacement("AB", 2) |
Three selections |
| Count ordered selections | perm(10, 3) |
720; Python 3.8+ |
| Count unordered selections | comb(10, 3) |
120; Python 3.8+ |
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