Permutations & Combinations

Permutations — when order matters

The number of ways to arrange k items chosen from n distinct items, where order matters:

P(n, k) = n · (n−1) · (n−2) · ... · (n−k+1) = n! / (n−k)!

For all of n items arranged: P(n, n) = n!.

Example — first, second, third place in a race with 8 runners. P(8, 3) = 8 · 7 · 6 = 336 ways. (Order matters because gold, silver, and bronze are different positions.)

Combinations — when order doesn't matter

The number of ways to choose k items from n distinct items, where order doesn't matter:

C(n, k) = n! / (k! · (n−k)!) = P(n, k) / k!

The "n choose k" notation. Read as "n choose k." Also written ⁿCₖ or (n, k) in superscript form.

Example — choose 5 cards from a 52-card deck. C(52, 5) = 52! / (5! · 47!) = 2,598,960 hands.

Permutation vs Combination — when to use which

Question	Order matters?	Use
"How many 4-digit PINs?"	Yes (1234 ≠ 4321)	10⁴ (with repetition) or P(10, 4) (without)
"How many ways to seat 8 people?"	Yes	P(8, 8) = 8! = 40,320
"How many 5-card poker hands?"	No (hand is a set)	C(52, 5) = 2,598,960
"How many committees of 3 from 12 people?"	No	C(12, 3) = 220
"First, second, third place from 10 racers?"	Yes	P(10, 3) = 720
"How many ways to pick 2 colors from 8?"	No	C(8, 2) = 28

Rule of thumb — if rearranging a chosen set gives a "different" outcome, use permutation; if it's the "same," use combination.

Worked examples

Example 1 — Powerball jackpot odds

5 numbers chosen from 69 (no repetition, no order) + 1 number from 26.

Main: C(69, 5) = 69!/(5! · 64!) = 11,238,513
Powerball: 26
Total: 11,238,513 × 26 = 292,201,338

Buying one ticket gives 1/292,201,338 chance of jackpot — about one in 290 million. You're more likely to be struck by lightning twice than win.

Example 2 — Birthday problem

23 people in a room. Probability that at least two share a birthday?

It's easier to compute the complement — probability all 23 birthdays are distinct.

P(all distinct) = (365/365) · (364/365) · (363/365) · ... · (343/365)
                = P(365, 23) / 365²³
                ≈ 0.4927

P(at least two shared) ≈ 1 − 0.4927 = 0.5073 ≈ 50.7%. Counterintuitive — only 23 people for >50% chance of a shared birthday.

Example 3 — Number of poker hands

5 cards from 52. C(52, 5) = 2,598,960. Probability of a specific hand type:

Hand	Number of hands	Probability
Royal flush	4	0.000154%
Straight flush	36	0.00139%
Four of a kind	624	0.024%
Full house	3,744	0.144%
Flush	5,108	0.197%
Straight	10,200	0.392%
Three of a kind	54,912	2.11%
Two pair	123,552	4.75%
One pair	1,098,240	42.3%
High card	1,302,540	50.1%

JavaScript — counting

// Factorial
function factorial(n) {
  let result = 1;
  for (let i = 2; i <= n; i++) result *= i;
  return result;
}

// Permutations P(n, k)
function permutations(n, k) {
  if (k > n) return 0;
  let result = 1;
  for (let i = 0; i < k; i++) result *= (n - i);
  return result;
}

// Combinations C(n, k)
function combinations(n, k) {
  if (k > n) return 0;
  if (k > n - k) k = n - k;  // symmetry: C(n, k) = C(n, n-k)
  let result = 1;
  for (let i = 0; i < k; i++) {
    result = result * (n - i) / (i + 1);
  }
  return Math.round(result);
}

// Examples
console.log(permutations(8, 3));   // 336
console.log(combinations(52, 5));  // 2598960
console.log(combinations(69, 5));  // 11238513

// Birthday problem
function birthdayProblem(people) {
  let pAllDifferent = 1;
  for (let i = 0; i < people; i++) {
    pAllDifferent *= (365 - i) / 365;
  }
  return 1 - pAllDifferent;
}
console.log(birthdayProblem(23).toFixed(3));  // 0.507
console.log(birthdayProblem(50).toFixed(3));  // 0.970

Variations and special cases

Problem	Formula
Permutations of n distinct items	n!
Permutations of k from n (no rep)	n! / (n − k)!
Permutations with repetition (k slots, n choices)	n^k
Permutations with identical groups	n! / (k₁! · k₂! · ... · kₘ!)
Combinations of k from n (no rep)	C(n, k) = n! / (k!(n−k)!)
Combinations with repetition	C(n+k−1, k)
Subsets (any size)	2ⁿ
Pairs of subsets	3ⁿ

Where this matters

• Probability. Lottery odds, card games, dice combinations, the birthday problem — all use combinations and permutations.
• Cryptography. Password space size — n^k for character × position. Key strength is measured in number of possible keys.
• Genetics. Number of possible genotypes from given alleles, distribution of traits in offspring.
• Statistics. Sampling without replacement, hypergeometric distribution, multinomial expansions.
• Computer science. Counting algorithm complexity (number of permutations to try, number of subsets to consider). Brute-force enumeration uses these counts directly.
• Operations research. Scheduling, assignment problems, traveling salesman (n! tours). Combinatorial optimization is built on these counts.
• Combinatorial design. Latin squares, balanced incomplete block designs, error-correcting codes.

Common mistakes

• Confusing permutations with combinations. The single most common error. "Order matters" is the test. If swapping two items gives a "different" answer, use permutation. If "same," use combination.
• Forgetting whether repetition is allowed. "Choose 4 letters from 26" — without repetition (P(26, 4) or C(26, 4)) or with (26⁴)? The two answers differ by orders of magnitude.
• Computing factorials directly for large n. 100! is too big to fit in any standard integer. Use the iterative product formula or BigInt; cancel before multiplying when possible.
• Forgetting 0! = 1. Both C(n, 0) = 1 and C(n, n) = 1 by convention. The empty arrangement counts as one valid arrangement.
• Mixing up identical objects. Permutations of MISSISSIPPI (1 M, 4 I, 4 S, 2 P) is 11! / (1!·4!·4!·2!) = 34,650, not 11!. Identical letters can't be told apart.
• Confusing P(n, k) with permutation P(X) of X. The function name is overloaded; check context.