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.