Binomial coefficient C(n,k) = n! / (k!(n−k)!)

The formula

$\binom{n}{k} = \dfrac{n!}{k! (n - k)!}$

Read as ”$n$ choose $k$”. The number of combinations of $k$ items chosen from $n$ (no order, no repetition).

Special cases

• $\binom{n}{0} = 1$ (one way to choose nothing)
• $\binom{n}{n} = 1$ (one way to choose everything)
• $\binom{n}{1} = n$
• $\binom{n}{2} = \dfrac{n(n-1)}{2}$

Symmetry

$\binom{n}{k} = \binom{n}{n - k}$

Ex: $\binom{10}{3} = \binom{10}{7}$. Choosing 3 is the same as leaving out 7.

Pascal’s triangle

Each number is the sum of the two above:

$\binom{n+1}{k+1} = \binom{n}{k} + \binom{n}{k+1}$

n=0:              1
n=1:            1   1
n=2:          1   2   1
n=3:        1   3   3   1
n=4:      1   4   6   4   1
n=5:    1   5  10  10   5   1

Application: binomial theorem

$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

For $n = 2$: $(a+b)^2 = a^2 + 2ab + b^2$ (coefficients 1, 2, 1).

For $n = 3$: $(a+b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3$ (coefficients 1, 3, 3, 1).

Application: combinations in probability

Number of 5-card hands from 52:

$\binom{52}{5} = \dfrac{52!}{5! \, 47!} = 2{,}598{,}960$

Common mistakes

1. Confusing combination and arrangement (arrangement = with order).
2. Forgetting the $(n - k)!$ in the denominator.
3. Computing huge factorials instead of simplifying (e.g., $10!/8! = 10 \times 9$).

?À toi de jouer

C(6, 2) = ?

A.15B.12C.30D.36

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