Permutations & Combinations Calculator

What's the difference between Permutations and Combinations?

Permutation: arrangement where ORDER matters. 'How many ways to arrange 3 books from 10?' nPr = n! / (n-r)! = 10! / 7! = 10×9×8 = 720. Combination: selection where order does NOT matter. 'How many ways to choose 3 books from 10?' nCr = n! / (r!(n-r)!) = 10! / (3!×7!) = 720/6 = 120. Memory aid: Combination lock → actually a Permutation lock (order matters for the code!). A real combination (selection) example: choosing lottery numbers. The ratio nPr = nCr × r! shows that permutations equal combinations multiplied by the number of ways to arrange each selected group, which is exactly why permutations always count at least as many outcomes as combinations for the same n and r.

What should I know about Pascal's Triangle and nCr?

Pascal's Triangle contains all nCr values. Row n, position r (starting from 0): nCr. Row 4: 1, 4, 6, 4, 1 = 4C0, 4C1, 4C2, 4C3, 4C4. Each entry = sum of two above. Key properties: nC0 = 1 (one way to choose nothing). nCn = 1 (one way to choose everything). nC1 = n. nCr = nC(n-r) (choosing r is same as excluding n-r). Symmetry: Pascal's Triangle is symmetric. Binomial theorem: (a+b)ⁿ = Σ nCr × aⁿ⁻ʳ × bʳ. The coefficients are the nCr values — this is why they are called binomial coefficients.

What should I know about Practical Applications?

Lottery (UK National Lottery): choose 6 from 59 numbers. 59C6 = 45,057,474. One in 45 million chance of winning the jackpot. Password security: 8-character password from 62 characters (a-z, A-Z, 0-9). Permutations with repetition: 62⁸ = 2.18×10¹⁴. With 10 billion password attempts per second: 6 hours to crack. 12 characters: 62¹² = 3.23×10²¹ → infeasible by brute force. Card games: 52-card deck. 5-card poker hand: 52C5 = 2,598,960. Royal flush: 4 ways (one per suit). Probability: 4/2,598,960 = roughly 1 in 650,000, showing just how rare a royal flush genuinely is even across a large number of hands dealt.

What do I need to know about Factorials?

n! = n × (n-1) × (n-2) × ... × 2 × 1. 0! = 1 (by definition — one way to arrange zero items). Stirling's approximation for large n: n! ≈ √(2πn) × (n/e)ⁿ. Factorials grow incredibly fast: 10! = 3,628,800. 20! = 2.43×10¹⁸. 52! = 8.07×10⁶⁷ (more than the number of atoms in Earth). This is why there are more possible orderings of a shuffled deck of cards than there are atoms on Earth — every properly shuffled deck has almost certainly never been in that exact order before.

What should I know about Permutations and Combinations with Repetition?

The standard nPr and nCr formulas assume each item can only be chosen once. When repetition is allowed, the formulas change: permutations with repetition (choosing r items from n options, order matters, repeats allowed) = nʳ — for a 4-digit PIN using digits 0-9, that's 10⁴=10,000 possible codes. Combinations with repetition (order doesn't matter, repeats allowed) uses the 'stars and bars' formula C(n+r−1, r) — for choosing 3 scoops of ice cream from 5 flavours where you can repeat a flavour, that's C(5+3−1,3)=C(7,3)=35 possible combinations. Recognising which of these four cases applies (ordered/unordered, with/without repetition) is the single biggest source of confusion in combinatorics problems, and it's always worth explicitly asking 'can an item be chosen more than once?' and 'does the order of selection matter?' before picking a formula.

What do I need to know about the Birthday Paradox as a Combinatorics Example?

The famous 'birthday paradox' — that in a room of just 23 people, there's over a 50% chance two share a birthday — is a direct application of combination counting. Rather than calculating the probability of a match directly, it's easier to calculate the probability of no match and subtract from 1. With 23 people and 365 possible birthdays, the number of ways to assign all-different birthdays is a permutation: 365×364×363×...×343 (23 terms), divided by the total possible assignments 365²³. That ratio comes out to about 0.493, so P(at least one match) = 1−0.493 ≈ 0.507. This result feels counterintuitive because people instinctively think about their own specific birthday matching someone else's, rather than the far larger number of possible pairs (C(23,2)=253 pairs) that could each potentially match — a useful illustration of why combination counts grow much faster than people expect.

What do I need to know about Permutations and Combinations with Constraints?

Many real counting problems add constraints beyond a plain "choose k from n," and these require adjusting the basic formulas rather than abandoning them. If specific items must always sit together in an arrangement (like a couple who must sit next to each other), treat that group as a single combined unit first, arrange the reduced set, then multiply by the number of ways to arrange the items within that unit. If specific items must never be adjacent, it's often easier to count all arrangements, then subtract the arrangements where they are adjacent, rather than counting the valid arrangements directly. If a selection must include at least one item from a specific subgroup, calculating "total combinations minus combinations with none from that subgroup" is typically far simpler than trying to directly count every valid case. Recognizing which of these indirect strategies fits a given constraint is usually the real skill being tested, more so than the base nCr or nPr formula itself.

Permutations vs Combinations: Side-by-Side Comparison

Notice that nCr is always nPr divided by r! — combinations simply remove the duplicate counts created by the different orderings of the same r items.

Worked Example: Odds of Winning a 6-from-59 Lottery

UK Lotto-style draws typically ask players to pick 6 numbers from a pool of 59, with no regard to the order they're drawn in — a combinations problem. The number of possible combinations is 59C6 = 59! / (6! × 53!). Working it through: the numerator product for the first 6 terms of 59! down to 54 is 59 × 58 × 57 × 56 × 55 × 54 = 32,441,381,280. Dividing by 6! = 720 gives 32,441,381,280 ÷ 720 = 45,057,474. So there are approximately 45,057,474 equally likely combinations, meaning a single ticket has roughly a 1-in-45-million chance of matching all 6 numbers — a concrete illustration of how fast combination counts grow, since adding just a few more numbers to the pool or balls to pick multiplies the total dramatically rather than adding to it.

Worked Example: Arrangements of a Word with Repeated Letters

Counting distinct arrangements of all the letters in a word with repeats uses the formula n! divided by the factorial of each letter's repeat count. The word MISSISSIPPI has 11 letters: M appears once, I appears 4 times, S appears 4 times, and P appears 2 times. The number of distinct arrangements is 11! / (1! × 4! × 4! × 2!). 11! = 39,916,800. The denominator is 1 × 24 × 24 × 2 = 1,152. Dividing: 39,916,800 ÷ 1,152 = 34,650 distinct arrangements — vastly fewer than the 39,916,800 arrangements there would be if all 11 letters were different, because swapping two identical letters with each other produces an arrangement that looks exactly the same and must only be counted once.

Worked Example: Combinations with Repetition (Choosing Ice Cream Scoops)

When choices can repeat and order doesn't matter — such as picking 3 scoops of ice cream from 5 available flavours, where you could pick the same flavour more than once — the relevant formula is 'combinations with repetition' (also called 'stars and bars'): C(n + r − 1, r), where n is the number of types available and r is the number of selections being made. With n = 5 flavours and r = 3 scoops: C(5 + 3 − 1, 3) = C(7, 3) = 7! / (3! × 4!) = 5040 / (6 × 24) = 5040 / 144 = 35. So there are 35 distinct ways to choose 3 scoops from 5 flavours when repeats are allowed and order doesn't matter (two scoops of chocolate and one of vanilla counts the same regardless of which was scooped first) — noticeably more than the 10 combinations (5C3) there would be if each flavour could only be chosen once.