Permutations & Combinations

In Phase 1 you learned to count step by step: if a choice has a options and the next has b options, the two together have a × b outcomes. That idea powers everything here. This phase asks two of the most common counting questions you'll ever meet — and the only thing separating them is a question you have to learn to ask out loud.

The two questions

Almost every "how many ways…" problem is one of these:

• Order matters. Gold, silver, bronze are different. Seat 1 and seat 2 are different. Counting these is permutations.
• Order doesn't matter. A handful of pizza toppings is the same handful in any order. A committee is the same committee no matter who you name first. Counting these is combinations.

Same raw ingredients — pick r things from n — but two very different counts. Before any formula, build the reflex: does swapping two of my chosen things change the answer? If yes, order matters. If no, it doesn't.

Factorial: counting orderings

Start with the simplest "order matters": arrange all of your items.

How many ways can 3 books sit on a shelf? Pick the leftmost: 3 choices. The next: 2 books remain. The last spot: 1 book left. By the multiplication principle that's 3 × 2 × 1 = 6.

That product — multiply every whole number from n down to 1 — is the factorial, written n!:

n! = n × (n − 1) × (n − 2) × … × 1

So 5! = 5 × 4 × 3 × 2 × 1 = 120. Factorials grow ferociously fast: 10! is already over three million.

One value surprises people: 0! = 1. Read it as "how many ways are there to arrange nothing?" Exactly one — the empty arrangement. It isn't a typo or a special case bolted on; it's the value that keeps the formulas below from breaking, as you'll see in a moment.

Permutations: arranging a few of many

Often you arrange a few items, not all of them.

Eight runners, and you want the top three: gold, silver, bronze. Order absolutely matters (gold is not bronze). Walk it like Phase 1:

• Gold: 8 runners could win it.
• Silver: 7 remain.
• Bronze: 6 remain.

That's 8 × 7 × 6 = 336 possible podiums.

Notice you multiplied only 3 of the factors of 8!, then stopped. The formula captures exactly that "stop early":

nPr = n! / (n − r)!

The (n − r)! in the denominator cancels off the factors you didn't use. Check it: 8! / 5! = (8 × 7 × 6 × 5!) / 5! = 8 × 7 × 6 = 336. The 5! divides clean away, leaving the three factors you actually counted.

Here n = 8 (the pool), r = 3 (the spots), and the order of those 3 is part of what makes each podium distinct.

Combinations: choosing without order

Now change one word in the question.

Eight pizza toppings, and you want three of them. Mushroom-pepper-onion is the same pizza as onion-mushroom-pepper. Order doesn't matter, so we should count fewer than 336 — because we decided a bunch of those 336 were duplicates.

How many duplicates? Any single group of 3 toppings can be listed in 3! = 6 orders, and the permutation count treated all 6 as separate. So we overcounted every real selection by a factor of 3!. Divide it back out:

336 / 6 = 56

That's the combination count, and the formula bakes in that division:

nCr = n! / (r! (n − r)!)

It's nPr with one extra r! in the denominator — the exact factor by which order inflated the count. Same eight things, same three picks, but 56 instead of 336.

See it run

Python has all three built in. Run this:

import math
print(math.factorial(5))   # 120
print(math.perm(8, 3))     # ordered: 8*7*6 = 336
print(math.comb(8, 3))     # unordered: 56

What just happened: math.factorial(5) printed 120 — the orderings of 5 items. math.perm(8, 3) printed 336 — the ordered top-3 podiums (8 × 7 × 6). math.comb(8, 3) printed 56 — the unordered 3-topping selections. The permutation count is larger than the combination count because order multiplies in extra arrangements: every one of those 56 selections shows up 3! = 6 times when order is tracked, and 56 × 6 = 336.

The deciding question, side by side

Hold the contrast in your head — the numbers are identical and the answers are not:

Both pick 3 from 8. The committee is six times smaller for one reason: a committee of {Ana, Ben, Cleo} is one committee, but as a ranking it's six different podiums. Decide order or not first; the formula is the easy part.

For builders

This shows up the moment you count possibilities in code or size a problem:

• Choosing a subset → combinations (nCr). Feature flags turned on, cards dealt from a deck, which sensors to sample, which test cases to bundle. The selection is a set; order is meaningless.
• Arranging or ranking → permutations (nPr). Tournament seedings, leaderboard top-N, the order tasks run in, password/PIN spaces where position matters.

Gut check before you reach for a library: if reordering your answer gives you "the same thing," you want comb. If reordering gives you "a different thing," you want perm. Both live in Python's math module, so you rarely write the factorials by hand.

⚠️ The classic mistake: using permutations when order doesn't matter. You'll overcount by exactly a factor of r!. That isn't a coincidence — it is the r! sitting in the denominator of nCr. If a count comes out suspiciously large (your 56 committees ballooned to 336), ask whether you accidentally counted the same group in every possible order.

Recap

• n! counts the orderings of n distinct items; 0! = 1.
• Permutations (nPr = n! / (n − r)!) count arrangements — order matters.
• Combinations (nCr = n! / (r! (n − r)!)) count selections — order doesn't.
• The difference between them is the factor r!, the number of ways to reorder your r picks.
• Always ask the deciding question first: does order matter? Everything else follows.

This whole machine is built from the multiplication principle and the idea of distinct items — if "distinct" feels slippery, Sets, relations, and functions makes the notion of a set (order-free, no repeats) precise. And if the formulas still feel like spells, Why math isn't your enemy is worth a detour.

Quick check before moving on:

[
  {
    "q": "What does n! (n factorial) count?",
    "choices": [
      "The number of ways to arrange n distinct items in order",
      "The number of ways to choose some of n items, ignoring order",
      "The sum of all whole numbers from 1 to n",
      "The number of subsets of an n-item set"
    ],
    "answer": 0,
    "explain": "n! = n × (n−1) × … × 1 multiplies the choices at each position, which is exactly the count of orderings of n distinct items."
  },
  {
    "q": "You're choosing a 3-person committee from 10 people. Permutation or combination?",
    "choices": [
      "Permutation, because you pick people one at a time",
      "Combination, because the committee is the same regardless of the order you name people",
      "Permutation, because 3 is smaller than 10",
      "Neither — it's a plain factorial"
    ],
    "answer": 1,
    "explain": "Reordering the committee members gives the same committee, so order doesn't matter. That's a combination (nCr)."
  },
  {
    "q": "How many ways can you choose 2 toppings from 4 (order doesn't matter)?",
    "choices": [
      "12",
      "8",
      "6",
      "4"
    ],
    "answer": 2,
    "explain": "nCr = 4! / (2! · 2!) = 24 / 4 = 6. (As a permutation it would be 4 × 3 = 12, which double-counts each pair.)"
  }
]

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