Permutation & Combination Calculator

Calculate P(n,r) = n!/(n−r)! for ordered arrangements and C(n,r) = n!/(r!(n−r)!) for unordered selections. Switch to Permutation & Combination tab for instant nPr and nCr results.

Option 1 — Favorable / Total Outcomes

OR

Option 2 — Two Events P(A) and P(B)

Guides & Reference

How It Works

Permutation P(n,r) — ordered arrangementsPasswords, race placements, seating, any order-dependent selection.

Switch to Permutation & Combination tab. Enter n (total items) and r (items chosen). P(n,r) = n!/(n−r)! = n×(n−1)×...×(n−r+1). This is r multiplications — efficient even for large n. Example: P(10,3) = 10×9×8 = 720. How many 3-letter passwords from 10 letters (no repeats, order matters).

P(n,r) = n!/(n−r)! = n×(n−1)×...×(n−r+1)P(10,3)=720 | P(52,5)=311,875,200 | P(8,8)=40,320
Combination C(n,r) — unordered selectionsCommittees, card hands, lottery, team formation, any order-independent selection.

C(n,r) = n!/(r!(n−r)!) = P(n,r)/r!. Dividing by r! removes the r! orderings of the same selection. Example: C(10,3) = P(10,3)/3! = 720/6 = 120. How many 3-person committees from 10 people (order doesn't matter — [A,B,C] = [C,A,B]).

C(n,r) = P(n,r) / r! = n! / (r!(n−r)!)C(10,3)=120 | C(52,5)=2,598,960 | C(52,13)=635,013,559,600
Pascal's Triangle and C(n,r)Understanding the pattern and computing combinations by hand.

Pascal's Triangle: each entry = sum of the two above. Row n gives all C(n,0), C(n,1), ..., C(n,n). Row 5: 1,5,10,10,5,1 → C(5,0)=1, C(5,1)=5, C(5,2)=10, C(5,3)=10, C(5,4)=5, C(5,5)=1. Symmetry: C(n,r)=C(n,n−r). Binomial theorem: (a+b)^n = Σ C(n,k)×a^k×b^(n−k).

C(n,r) = C(n−1,r−1) + C(n−1,r) (Pascal's rule)C(5,2)=10 | Row 4: 1,4,6,4,1 | C(4,2)=6
Binomial Distribution connectionProbability of k successes in n trials — directly uses C(n,k).

P(X=k) = C(n,k) × p^k × (1−p)^(n−k). The combination C(n,k) counts how many of the n! orderings give exactly k successes in n positions. Switch to Binomial Distribution tab and enter n, k, and p. Example: P(3 heads in 10 flips) = C(10,3)×0.5^10 = 120/1024 ≈ 11.72%.

P(X=k) = C(n,k) × p^k × (1−p)^(n−k)n=10, k=3, p=0.5: C(10,3)=120, P≈11.72%
Large combinations — no overflowCryptography key space, genome sequencing, large-scale combinatorics.

The calculator uses integer arithmetic to avoid floating-point overflow. C(52,26) = 495,918,532,948,104 (about 500 trillion). C(100,50) has 29 digits. Results display exactly with digit count. For very large n and r, the computation may take a moment but remains exact.

BigInt ensures exact results for large C(n,r) and P(n,r)C(52,26) = 495,918,532,948,104 (exact, 15 digits)

Quick Reference

Verify these in the calculator above.

Permutation

P(10,3)

720

Permutation

P(8,8)

40,320

Combination

C(10,3)

120

Poker hands

C(52,5)

2,598,960

Combination

C(6,2)

15

Symmetry

C(n,r) = C(n,?)

C(n,n−r)

Bridge hand

C(52,13)

635 billion+

All arrange

P(6,6) = 6!

720

Tips & Shortcuts

Remember C(n,r) = C(n, n−r). C(52,49) = C(52,3) = 22,100 — much faster to compute C(52,3) directly.

For "at least" problems: P(at least 1) = 1 − P(none) = 1 − C(n,0)×p^0×(1−p)^n. Use the Binomial tab for cumulative probabilities.

Permutation of all n items: P(n,n) = n!. The number of ways to arrange an entire set. For 8 people in a line: 8! = 40,320.

Combination C(n,1) = n always. Choosing 1 item from n: n ways. C(n,n) = 1 always. Choosing all n items: only 1 way.

Use the Permutation & Combination tab for direct nPr and nCr computation — the Basic Probability tab is for P(event) = favorable/total.

Common Mistakes

Using permutation when combination is needed

For committees, groups, or any selection where order doesn't matter: use C(n,r). Using P(n,r) overcounts by r! — each selection counted r! times instead of once.

Forgetting to subtract r from n for permutation

P(n,r) = n×(n−1)×...×(n−r+1) — exactly r multiplications. P(10,3) = 10×9×8 = 720, not 10×9×8×7×...×1 = 10!.

Entering r > n

C(n,r) and P(n,r) are undefined for r > n — you cannot choose more items than exist. The calculator shows an error. C(5,6) has no meaning.

Confusing C(n,r) with n/r

C(10,3) = 120, not 10/3 = 3.33. The combination formula involves factorials, not simple division. C(n,1) = n (only then is it equal to n/1).

Applying factorial directly for combinations

C(10,3) ≠ 10! / 3! = 3,628,800/6 = 604,800. The correct formula divides by BOTH r! AND (n−r)!: 10!/(3!×7!) = 3,628,800/(6×5040) = 120.

Frequently Asked Questions

Permutation P(n,r): order matters — ABC and BAC are different. Combination C(n,r): order does not matter — ABC and BAC are the same selection. P(n,r) = n!/(n−r)!. C(n,r) = n!/(r!(n−r)!). Always: P(n,r) ≥ C(n,r).

P(5,3) = 5!/(5−3)! = 5!/2! = 120/2 = 60. This counts ordered selections: how many ways to arrange 3 items from 5 (ABC≠BAC).

C(10,3) = 10!/(3!×7!) = (10×9×8)/(3×2×1) = 720/6 = 120. This counts unordered selections: how many groups of 3 from 10 (order doesn't matter).

Permutation: passwords, race placements, seating arrangements, ordered codes. Combination: committees, card hands, lottery picks, team selections. Rule: if swapping items gives a different valid outcome, use permutation. If not, use combination.

Yes — choosing r items from n is the same as leaving n−r items out. C(10,3) = C(10,7) = 120. Choosing a team of 3 from 10 is equivalent to excluding 7. This symmetry halves computation for large combinations.

8! = 40320. This is P(8,8) — a permutation of all items. For 3 of 8 people in 3 specific seats: P(8,3) = 8×7×6 = 336.

C(52,5) = 52!/(5!×47!) = 2,598,960. There are almost 2.6 million possible five-card hands from a standard 52-card deck. The probability of any specific hand is 1/2,598,960 ≈ 0.0000385%.

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