Permutations and combinations

Permutations and combinations are counting rules: a permutation counts the ordered arrangements of r items chosen from n, and a combination counts the unordered selections.

Last updated: 07 Oct, 2026 · SciPy 1.18

Every probability in Probability basics was a count of outcomes. With a die you can list them, but with 6 chocolates, 52 cards or 100 coin tosses you need a formula for the count. Permutations and combinations are those formulas.

Counting ordered lists of chocolates

The video's school trip goes to a chocolate factory that makes 6 kinds: Dairy Milk, 5 Star, Milky Bar, Eclairs, Gems and Silk. A student's assignment is to write down the first three chocolates they see, in the order they see them.

The first name can be any of 6 chocolates. Once it is written, 5 are left for the second name, then 4 for the third. So there are 6 × 5 × 4 = 120 possible lists. "Dairy Milk, Gems, Milky Bar" and "Milky Bar, Gems, Dairy Milk" use the same chocolates but are different lists. Each ordered list is a permutation.

Here n = 6 is the number of chocolates and r = 3 is the number of names written. n! (n factorial) is n × (n − 1) × … × 1, and the (n − r)! cancels the part of 6! that is never used.

Counting unordered groups of chocolates

In a combination only the group matters, not the order. Dairy Milk, Gems and Eclairs is one combination however the three are arranged; swapping them around does not make a new one.

The two counts are tied together. Each group of three can be put in order in 3! = 3 × 2 × 1 = 6 ways, so each group appears 6 times among the 120 lists. Dividing removes the repeats: 120 ÷ 6 = 20. In general nCr = nPr ÷ r!.

The question decides which to use. "The first three chocolates you see, in order" is a permutation. "Which three chocolates go into a gift box" is a combination. Both formulas assume n different items and no item used twice. When repeats are allowed, there are nʳ ordered lists (6³ = 216) and C(n + r − 1, r) unordered ones (C(8, 3) = 56).

Counting with math and itertools

math.perm and math.comb give the counts directly; itertools builds the lists themselves, so the counts can be checked one by one.

import math
from itertools import permutations, combinations, combinations_with_replacement

chocolates = ["Dairy Milk", "5 Star", "Milky Bar", "Eclairs", "Gems", "Silk"]
print("6P3 =", math.perm(6, 3), "  6C3 =", math.comb(6, 3), "  3! =", math.factorial(3))

lists = list(permutations(chocolates, 3))       # ordered
groups = list(combinations(chocolates, 3))      # unordered
print("ordered lists:", len(lists), "  groups:", len(groups), "  120 / 6 =", len(lists) // 6)
print("first two lists:", lists[:2])

one = {"Dairy Milk", "Gems", "Eclairs"}
print("lists using exactly that group:", sum(1 for l in lists if set(l) == one))

print("with repeats, ordered:", 6 ** 3, "  unordered:", len(list(combinations_with_replacement(chocolates, 3))))
print("P(a random group of 3 is that group) =", 1 / math.comb(6, 3))
print("5-card hands from 52 cards:", math.comb(52, 5))

6P3 = 120   6C3 = 20   3! = 6
ordered lists: 120   groups: 20   120 / 6 = 20
first two lists: [('Dairy Milk', '5 Star', 'Milky Bar'), ('Dairy Milk', '5 Star', 'Eclairs')]
lists using exactly that group: 6
with repeats, ordered: 216   unordered: 56
P(a random group of 3 is that group) = 0.05
5-card hands from 52 cards: 2598960

What the chocolate counts show

• 6P3 = 120 and 6C3 = 20: the two counts worked out on the board.
• itertools builds 120 lists and 20 groups, and 120 ÷ 6 = 20 matches.
• The group {Dairy Milk, Gems, Eclairs} appears in 6 of the lists: its 3! orderings.
• Repeats change the counts: 216 ordered lists and 56 groups when a chocolate may appear more than once.
• A random group is that one group with probability 0.05, 1 out of 20, and there are 2,598,960 five-card poker hands.

Permutation vs combination

Where you use permutations and combinations

• The binomial distribution: C(n, k) counts the ways to get k heads in n tosses, the first step of the Binomial distribution formula.
• Odds of a draw: lottery tickets, card hands and passwords are counts of combinations or permutations.
• Model search: choosing 3 of 10 features gives C(10, 3) = 120 subsets to try.

Related

• Previous: Conditional probability and Bayes' theorem
• Next: Random variables and probability distributions