Mathematics · Stochastics / Combinatorics

Binomial Coefficient (n choose k)

The binomial coefficient n choose k counts how many k-element subsets can be selected from n objects, without regard to order.

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Formula

(k n) = \frac{n !}{k ! ( n - k )!}

LaTeX: \binom{n}{k} = \frac{n!}{k!\,(n-k)!}

Dimensionless (counts)

Variables & units – Binomial Coefficient (n choose k)

Symbol	Meaning	Unit
n	Number of available objects	dimensionless
k	Number of selected objects (0 ≤ k ≤ n)	dimensionless
n!	Factorial n! = n·(n−1)·...·1, with 0! = 1	dimensionless

Derivation & background – Binomial Coefficient (n choose k)

The binomial coefficient counts combinations without repetition. If order matters, the permutation count nPk = n!/(n−k)! = (n choose k)·k! applies. In Pascal's triangle (Traité 1665, known in Asia centuries earlier) the building rule (n choose k) = (n−1 choose k−1) + (n−1 choose k) and the symmetry (n choose k) = (n choose n−k) hold. On calculators the function is called nCr. In the binomial distribution it counts the paths with k successes.

Exam blueprint

Validity range

Counts combinations without repetition: n distinct objects, choose k of them, order irrelevant, no replacement. With order nPk applies, with replacement other counting formulas.

Derivation steps

Count ordered first, then divide by the arrangements of the selection.

1Ordered selections: n·(n−1)·...·(n−k+1) = n!/(n−k)! possibilities.
2Every unordered selection was counted k! times; division gives n!/(k!(n−k)!).

Rearrangements

Permutation (with order)

n P k = \frac{n !}{( n - k )!} = (k n) \cdot k!

Podium places instead of teams: order matters.

Symmetry

(k n) = (n - k n)

Choosing k means leaving out n−k; saves computation.

Pascal rule

(k n) = (k - 1 n - 1) + (k n - 1)

Building principle of Pascal's triangle.

Task variant

How many teams of 3 can be formed from 10 people?

(10 choose 3) = (10·9·8)/(3·2·1) = 720/6 = 120 teams.

From 5 women and 4 men, 2 of each are to be chosen. How many possibilities are there?

(5 choose 2)·(4 choose 2) = 10·6 = 60 possibilities (partial selections multiply).

Common mistakes

Confusing combination and permutation.

Order irrelevant: nCk. Order matters: nPk = nCk·k!.

Computing n!/(k!(n−k)!) fully and risking overflow.

Cancel: (n choose k) = n·(n−1)·...·(n−k+1)/k!; large factorials cancel.

Assuming 0! = 0.

0! = 1 by definition; hence (n choose 0) = (n choose n) = 1.

Exam context

Combinatorics tasks (lottery, committees), path counts in the binomial distribution, hypergeometric models.

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Formula cluster

Counting and distributions

The counting tool behind the binomial distribution and many probability models.

P (X = k) = (k n) \cdot p^{k} \cdot (1 - p)^{n - k}

Binomial Distribution

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )} (P (B) > 0)

Conditional Probability

E (X) = μ = \sum_{i} x_{i} \cdot p_{i}

Expected Value

Worked example

Lottery 6 out of 49: (49 choose 6) = 49!/(6!·43!) = 13,983,816 possible tickets. Small example: (10 choose 3) = (10·9·8)/(3·2·1) = 120.

Applications

Lottery and selection tasks, path counts in Bernoulli chains (binomial distribution), quality control (samples), team compositions

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Which formula describes Binomial Coefficient (n choose k)?

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How do you rearrange (n über k) = n!/(k!·(n−k)!) for Permutation (with order)?

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Which common mistake happens with Binomial Coefficient (n choose k)?

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Scientific sources

[1]Pascal, B. (1665). Traité du triangle arithmétique. Paris.

Common notations & search queries

n über k FormelnCr Formeln!/(k!(n-k)!)Binomialkoeffizient berechnenKombinationen ohne WiederholungnPr Permutationbinomial coefficientLotto 6 aus 49 MöglichkeitenPascalsches Dreieck

Related formulas

P (X = k) = (k n) \cdot p^{k} \cdot (1 - p)^{n - k}

Binomial Distribution

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )} (P (B) > 0)

Conditional Probability

E (X) = μ = \sum_{i} x_{i} \cdot p_{i}

Expected Value

More Mathematics formulas

a^{2} + b^{2} = c^{2}

Pythagorean Theorem

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

Quadratic Formula

e^{i φ} = cos φ + i \cdot sin φ

Euler's Formula

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

Product Rule of Differentiation

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Frequently asked questions about Binomial Coefficient (n choose k)

How do you compute n choose k without a calculator?+

Use the reduced product form instead of full factorials: (n choose k) = n·(n−1)·...·(n−k+1)/k!, i.e. k descending factors from n, divided by k!. Example: (10 choose 3) = (10·9·8)/(3·2·1) = 720/6 = 120. Even faster with the symmetry (n choose k) = (n choose n−k): instead of (49 choose 43) compute (49 choose 6). Small values can be read directly from Pascal's triangle, where every entry is the sum of the two above. Boundary values by heart: (n choose 0) = 1, (n choose 1) = n, (n choose n) = 1. Computing full factorials like 49! produces huge numbers and rounding errors needlessly.

When do you use combinations, when permutations?+

The test question is: does order matter? If no (teams, lottery numbers, card hands), you count combinations: (n choose k) = n!/(k!(n−k)!). If yes (podium places, PIN digits, offices like chair and treasurer), you count permutations: nPk = n!/(n−k)! = (n choose k)·k!. The factor k! is exactly the number of arrangements of a fixed selection. Example with 10 people and 3 places: a team of 3 exists (10 choose 3) = 120 times; gold, silver and bronze however 10·9·8 = 720 times, because every group of three can be placed on the podium in 3! = 6 ways. For drawing with replacement and order, nᵏ applies instead.

Why is 0! = 1 and what does n choose 0 mean?+

0! = 1 is a definition, but a compelling one: the empty product has value 1 (just as the empty sum has value 0), and only this keeps the arithmetic rules consistent, e.g. n! = n·(n−1)! also for n = 1. Combinatorially (n choose 0) = 1 means: there is exactly one way to select nothing from n objects, namely the empty selection. Likewise (n choose n) = 1, selecting all works in only one way. With 0! = 0 the formula n!/(k!(n−k)!) would divide by 0 at the boundaries and collapse. The binomial distribution needs this too: P(X = 0) = (n choose 0)·p⁰·(1−p)ⁿ = (1−p)ⁿ works only with (n choose 0) = 1.

How likely is hitting the jackpot in a 6-out-of-49 lottery?+

There are (49 choose 6) = 49·48·47·46·45·44/6! = 13,983,816 possible draw outcomes, and exactly one matches your ticket. The probability is 1/13,983,816 ≈ 7.15·10⁻⁸, about 0.000007 %. For perspective: with one ticket per week you wait on average about 269,000 years for the jackpot. Smaller prize tiers are also computed with binomial coefficients: exactly 4 correct numbers have probability (6 choose 4)·(43 choose 2)/(49 choose 6) = 15·903/13,983,816 ≈ 0.097 %, because 4 must come from the 6 drawn and 2 from the 43 remaining numbers. This product pattern (hits times non-hits) is called the hypergeometric distribution.

What does the binomial coefficient have to do with the binomial distribution?+

It is its counting core. In a Bernoulli chain of n trials, every specific path with exactly k successes has probability pᵏ·(1−p)ⁿ⁻ᵏ. But there are many such paths: namely (n choose k) ways to distribute the k success positions among the n trials. Hence the formula P(X = k) = (n choose k)·pᵏ·(1−p)ⁿ⁻ᵏ. Example: 3 heads in 10 tosses of a fair coin: each path has (1/2)¹⁰ = 1/1024, there are (10 choose 3) = 120 paths, together 120/1024 ≈ 11.7 %. Whoever drops the binomial coefficient computes only the probability of ONE specific pattern, e.g. heads in the first three tosses.

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How do you calculate with Binomial Coefficient (n choose k)?

Here is how to work through a typical Binomial Coefficient (n choose k) ((n über k) = n!/(k!·(n−k)!)) task step by step:

1
Task
How many teams of 3 can be formed from 10 people?
Solution path
(10 choose 3) = (10·9·8)/(3·2·1) = 720/6 = 120 teams.
2
Task
From 5 women and 4 men, 2 of each are to be chosen. How many possibilities are there?
Solution path
(5 choose 2)·(4 choose 2) = 10·6 = 60 possibilities (partial selections multiply).