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Category Statistics & Probability

Total items (n) Integer from 0 to 1000

Items chosen (k) Integer from 0 to n

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About

Miscounting combinations is a common source of error in probability, statistics, and experimental design. The binomial coefficient C(n, k) counts the number of ways to choose k items from n distinct items where order does not matter and no item is selected twice. A manual factorial calculation for n = 50 already produces a 65-digit number, making overflow inevitable on standard 64-bit floats. This calculator uses arbitrary-precision integer arithmetic (BigInt) to return exact results for inputs up to n = 1000. It also provides a step-by-step multiplicative expansion and a Pascal's Triangle fragment so you can visually verify the result.

Note: this tool computes combinations without repetition only. If your problem allows selecting the same item more than once, you need the multiset coefficient C(n + k − 1, k). The formula assumes 0 ≤ k ≤ n. By convention C(n, 0) = 1 and C(0, 0) = 1.

Formulas

The number of combinations without repetition (binomial coefficient) is defined as:

C(n, k) = n!k! ⋅ (n − k)!

To avoid computing huge factorials, the multiplicative (product) form is used instead:

C(n, k) = k∏i = 1 n − k + ii

The symmetry property C(n, k) = C(n, n − k) is exploited: computation uses min(k, n − k) iterations. Pascal's recurrence relation provides an alternative verification path:

C(n, k) = C(n − 1, k − 1) + C(n − 1, k)

Where: n = total number of distinct items in the set. k = number of items chosen. n! = factorial of n (1 × 2 × … × n). By definition, 0! = 1.

Reference Data

n	C(n,1)	C(n,2)	C(n,3)	C(n,4)	C(n,5)	C(n,6)	C(n,7)	C(n,8)
5	5	10	10	5	1	-	-	-
6	6	15	20	15	6	1	-	-
7	7	21	35	35	21	7	1	-
8	8	28	56	70	56	28	8	1
10	10	45	120	210	252	210	120	45
13	13	78	286	715	1287	1716	1716	1287
20	20	190	1140	4845	15504	38760	77520	125970
26	26	325	2600	14950	65780	230230	657800	1562275
30	30	435	4060	27405	142506	593775	2035800	5852925
36	36	630	7140	58905	376992	1947792	8347680	30260340
45	45	990	14190	148995	1221759	8145060	45379620	215553195
49	49	1176	18424	211876	1906884	13983816	85900584	450978066
52	52	1326	22100	270725	2598960	20358520	133784560	752538150
60	60	1770	34220	487635	5461512	50063860	386206920	2558620845
100	100	4950	161700	3921225	75287520	1192052400	16007560800	186087894300

Frequently Asked Questions

This calculator accepts values of n up to 1000. It uses JavaScript BigInt arithmetic, which handles arbitrary-precision integers. The result for C(1000, 500) has 300 digits and is computed exactly. No floating-point approximation is involved.

The factorial definition requires computing n! which grows astronomically. For n = 170, the factorial already exceeds the IEEE 754 double-precision range (~1.7 × 10³⁰⁸). The multiplicative formula computes the product incrementally: multiply by (n − k + i) then divide by i at each step. Combined with BigInt, intermediate values stay manageable and exact.

Permutations count ordered arrangements: P(n, k) = n! ÷ (n − k)!. Combinations disregard order, so C(n, k) = P(n, k) ÷ k!. For example, choosing 5 cards from 52: P(52, 5) = 311,875,200 ordered hands, but C(52, 5) = 2,598,960 unordered hands.

There is exactly one way to choose zero items from any set: take nothing. Mathematically, C(n, 0) = n! ÷ (0! × n!) = 1. Similarly C(n, n) = 1 because there is only one way to take everything.

Most lotteries require selecting k numbers from n without repetition and without order. The 6/49 lottery has C(49, 6) = 13,983,816 possible outcomes, giving a jackpot probability of approximately 7.15 × 10⁻⁸. Powerball uses C(69, 5) × 26 = 292,201,338 combinations.

Yes. The calculator automatically uses min(k, n − k) as the loop bound. For C(1000, 998), it performs only 2 iterations instead of 998. This reduces time complexity from O(k) to O(min(k, n − k)).