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Algebra · Probability

Combinations Calculator

Calculate combinations C(n, r) using a high-precision arbitrary BigInt math engine with repetition modes and subset generation.

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Total Items (n)

Choose Items (r)

Choosing 3 items from a set of 10.

Total Combinations

120

There are exactly 120 unique subsets possible.

📂 Mathematical Substitution Steps

C(n,r) = n!r! (n - r)!

Formula: C(n, r) = n! / [ r! * (n - r)! ]

Substitute: C(10, 3) = 10! / [ 3! * (10 - 3)! ]

Factorials: 10! = 3628800

3! = 6

(7)! = 5040

Result: C(10, 3) = 120

Subsets Listing (120 total)

Generated list for sets n ≤ 15, r ≤ 5.

{ 1, 2, 3 }

{ 1, 2, 4 }

{ 1, 2, 5 }

{ 1, 2, 6 }

{ 1, 2, 7 }

{ 1, 2, 8 }

{ 1, 2, 9 }

{ 1, 2, 10 }

{ 1, 3, 4 }

{ 1, 3, 5 }

{ 1, 3, 6 }

{ 1, 3, 7 }

C(10, r) Variations

Choose (r)	Unique Ways
0	1
1	10
2	45
3	120
4	210
5	252

Pascal's Triangle Row 10

Pascal row coefficients map exactly to binomial expansion factors of C(n, r).

1104512021025221012045101

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Combinations vs Permutations

The fundamental core difference between Combinations and Permutations boils down to order.

In Permutations, the relative sequence or sequencing order matters (such as credentials passcode sequences). In Combinations, order is completely irrelevant, and only the unique elements of the chosen subset matter (like picking cards in a hand).

Combinations (nCr)

{A, B} equals {B, A}

Permutations (nPr)

(A, B) is distinct from (B, A)

Common FAQs

What does standard nCr signify?

nCr is standard mathematical shorthand for 'n choose r', representing total subset combinations possible from an overall pool.

Can 'r' be strictly greater than 'n'?

Under standard modes without repetition, choosing more items than are physically available yields 0. Under with-repetition mode, choosing larger amounts is valid since elements can be drawn multiple times.

How are combinations used in lottery odds?

To win a standard lottery requiring 6 correctly selected numbers from 49 total, we compute C(49, 6) which equals 13,983,816 possible single tickets, representing a 1 in ~14 million probability.

How Many Ways Can You Choose?

Combination or permutation — the only question

Both count selections; the difference is whether order matters:

Scenario	Order matters?	Count with
Pick 3 committee members from 10	No	Combinations: C(10,3) = 120
Award gold/silver/bronze among 10	Yes	Permutations: P(10,3) = 720
Choose 6 lottery numbers from 49	No	C(49,6) = 13,983,816
Form a 4-digit PIN	Yes (with repeats)	10^4 = 10,000
Pizza with any 3 of 8 toppings	No	C(8,3) = 56

The formula, demystified

C(n, k) = n! ÷ (k! × (n−k)!) — count all ordered arrangements, then divide out the orderings of the chosen group (k!) because order doesn't matter. Sanity anchors: C(n, 1) = n, C(n, n) = 1, and C(n, k) = C(n, n−k) — choosing 3 to include equals choosing 7 to exclude from 10. When a problem says “arrange”, “order” or “sequence”, you've left combinations for permutations.

From counting to probability

Most probability problems are two combination counts divided: favorable outcomes over total outcomes — the lottery odds above are exactly 1 ÷ C(49,6). The probability calculator packages that step, and descriptive statistics on actual data live in the statistics calculator.