What is the difference between combinations and permutations?

Combinations count the number of ways to choose r items from n when ORDER DOES NOT MATTER. Permutations count when ORDER MATTERS. Example: choosing 2 letters from A, B, C. - Combinations: {A,B}, {A,C}, {B,C} = 3 (AB same as BA) - Permutations: AB, BA, AC, CA, BC, CB = 6 (AB different from BA) Formulas: - Combinations: C(n,r) = n! / (r! x (n-r)!) = 3 - Permutations: P(n,r) = n! / (n-r)! = 6 - Relationship: P(n,r) = C(n,r) x r! (multiply by the r! ways to arrange the chosen items) When to use which: - Combinations: lottery numbers, card hands, committee selection, choosing pizza toppings - Permutations: race finishing order, PIN codes, password arrangements, seating order

How do you calculate C(n,r) step by step?

C(n,r) = n! / (r! x (n-r)!) For C(10,3): Step 1: n=10, r=3, n-r=7 Step 2: C(10,3) = 10! / (3! x 7!) Step 3: Expand: (10 x 9 x 8 x 7!) / (3! x 7!) = (10 x 9 x 8) / (3 x 2 x 1) Step 4: = 720 / 6 = 120 Tip for large numbers: cancel (n-r)! from numerator and denominator first, then compute only the top r terms: C(10,3) = (10 x 9 x 8) / (3 x 2 x 1) = 120 For C(52,5) poker hands: C(52,5) = (52 x 51 x 50 x 49 x 48) / (5 x 4 x 3 x 2 x 1) = 311,875,200 / 120 = 2,598,960

What is Pascal's Triangle and how does it relate to combinations?

Pascal's Triangle is a triangular array where each number equals the sum of the two numbers directly above it. Row n and position r (counting from 0) equals C(n,r). Row 0: 1 Row 1: 1 1 Row 2: 1 2 1 Row 3: 1 3 3 1 Row 4: 1 4 6 4 1 So C(4,2) = 6 (row 4, position 2). C(3,1) = 3 (row 3, position 1). Pascal's rule: C(n,r) = C(n-1,r-1) + C(n-1,r) Each value is the sum of the two values in the row above it. Properties of Pascal's Triangle: - Each row sums to 2^n: row 4 sums to 1+4+6+4+1=16=2^4 - Symmetrical: each row reads the same forwards and backwards - The triangle appears in the binomial theorem: (a+b)^n expansion coefficients are row n - Fibonacci numbers appear along the diagonals

What is the symmetry property of combinations?

C(n,r) = C(n, n-r). Choosing r items from n is the same as choosing which n-r items to EXCLUDE. Examples: - C(10,3) = C(10,7) = 120. Choosing 3 from 10 is equivalent to leaving out 7. - C(52,5) = C(52,47) = 2,598,960. Choosing a 5-card hand is the same as choosing the 47 cards not in the hand. - C(6,2) = C(6,4) = 15. Why this matters: - Always compute C(n,r) with r = min(r, n-r) to minimise computation - C(100,97) = C(100,3) = 161,700 — much easier to compute as C(100,3) - The symmetry is why Pascal's Triangle reads the same left to right as right to left

What are real-world examples of combinations?

Combinations appear whenever you select a group and the order does not matter. Lottery: UK National Lottery (6 from 59) = C(59,6) = 45,057,474 possible tickets. Powerball (5 from 69 + 1 from 26) = C(69,5) x 26 = 292,201,338 combinations. Cards: C(52,5) = 2,598,960 distinct 5-card poker hands. C(52,13) = 635,013,559,600 possible bridge hands. Teams and committees: choosing a 5-person team from 20 candidates = C(20,5) = 15,504 ways. Pizza toppings: choosing 3 toppings from a menu of 10 = C(10,3) = 120 combinations. Genetics: the number of ways to choose which 2 of 23 chromosome pairs to recombine = C(23,2) = 253. Cryptography: combinations are fundamental to counting keys, hash collisions, and birthday attack probabilities.

Combination Calculator (nCr) — C(n,r) = n! / (r! x (n-r)!)

The Combination Calculator computes C(n,r) — the number of ways to choose r items from n items when order does not matter. Uses exact BigInt arithmetic so results are always precise even for large values like C(52,5)=2,598,960 or C(100,50). Shows the step-by-step factorial cancellation, permutations...

C(n, r) = ?

CHOOSE r ITEMS FROM n ITEMS (ORDER DOES NOT MATTER)

n (total)

r (choose)

)

C(10, 3) = NUMBER OF COMBINATIONS

120

= 10! / (3! × 7!)

Permutations P(n,r)

720

order matters

Symmetry C(n, n-r)

120

C(10, 7)

Probability (1/nCr)

0.00833333

if chosen randomly

Pascal's rule check

120 ✓

C(n-1,r-1)+C(n-1,r)

STEP-BY-STEP

n = 10, r = 3

C(n, r) = n! / (r! × (n−r)!)

C(10, 3) = 10! / (3! × 7!)

= 3,628,800 / (6 × 5,040)

= 120

Symmetry check: C(10, 7) = 120 ✓

Pascal's rule: C(9,2) + C(9,3) = 36 + 84 = 120 ✓

C(n,0) = C(n,n)1

C(n,1) = C(n,n-1)10

C(n,2)45

Sum of row n1,024 (= 2ⁿ)

QUICK EXAMPLES

PASCAL'S TRIANGLE

Row n, position r (highlighted in black)

PROPERTIES OF C(10, 3)

Result120

Formula10! / (3! × 7!)

SymmetryC(10,3) = C(10,7)

Permutations P(n,r)720

P(n,r) / C(n,r)3! = 6

Sum of all C(n,k), k=0..n2^10 = 1,024

WHAT THIS MEANS

There are 120 ways to choose 3 items from 10 items when order does not matter.

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FORMULAS

COMBINATION & PERMUTATION FORMULAS

NAME	FORMULA	DESCRIPTION
Combination nCr	C(n,r) = n! / (r! x (n-r)!)	Number of ways to choose r from n, order irrelevant
Permutation nPr	P(n,r) = n! / (n-r)!	Number of ways to choose r from n, order matters
Relationship	P(n,r) = C(n,r) x r!	Permutations = combinations x arrangements of chosen items
Symmetry	C(n,r) = C(n, n-r)	Choosing r items same as choosing which n-r to exclude
Pascal's rule	C(n,r) = C(n-1,r-1) + C(n-1,r)	Each value is sum of the two above it in Pascal's Triangle
Row sum	Sum of C(n,k) for k=0 to n = 2^n	Total combinations across all r values equals 2 to the n
Special cases	C(n,0) = C(n,n) = 1	Choosing none or all gives exactly 1 way
Linear case	C(n,1) = C(n,n-1) = n	Choosing one item gives n options

HOW TO USE

1
Enter n — the total number of items in the set. For example, n=52 for a standard deck of cards.
2
Enter r — how many items you are choosing. Order does not matter: choosing items A, B, C is the same as B, C, A.
3
C(n,r) appears instantly in the black result box as an exact integer. For large n, BigInt arithmetic is used for exact results.
4
Check the related values: permutations P(n,r) = C(n,r) x r! (when order matters), symmetry C(n,r) = C(n,n-r), and the probability 1/C(n,r) of a single random selection.
5
The Pascal's Triangle diagram highlights your result at row n, position r. Each cell equals the sum of the two cells above it, which is Pascal's rule: C(n,r) = C(n-1,r-1) + C(n-1,r).

WORKED EXAMPLE

C(10,3): n=10, r=3. Cancel 7! top and bottom. (10x9x8)/(3x2x1) = 720/6 = 120. Symmetry: C(10,7)=120. Pascal: C(9,2)+C(9,3)=36+84=120. Permutations P(10,3)=120x6=720. Probability=1/120=0.00833. C(52,5): (52x51x50x49x48)/(5x4x3x2x1)=311875200/120=2598960 poker hands.

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Last updated: April 29, 2026 · Formula verified by EagleCalculator team · Eagle-eyed accuracy for every calculation.