= Binomial coefficient
{title2=$\binom{n}{k}$}
{wiki}
For <nonnegative integers> $n$ and $0\leq k\leq n$, the binomial coefficient
$$
\binom nk=\frac{n!}{k!(n-k)!}
$$
counts the $k$-element <subsets> of an $n$-element <set>. To obtain the formula, first choose an ordered list of $k$ different members in $n(n-1)\cdots(n-k+1)$ ways. Each <subset> appears in exactly $k!$ orders, so divide by that <factorial>. The cases $k=0$ and $k=n$ each have one selection, the <empty set> and the whole <set>. For integer $k$ outside $0\leq k\leq n$, the combinatorial convention is $\binom nk=0$.
For $n=5$, $k=2$, the calculation is $\binom52=5\cdot4/2=10$. The left-hand panel lists all ten <subsets> of $\{a,b,c,d,e\}$. Marking an included member with one and an excluded member with zero gives a <bijection> with length-five <binary strings> containing exactly two ones. More generally, this makes selection, binary encoding, and coefficients in a <binomial theorem> three descriptions of the same count.
\Image[/binomial-subsets-and-triangle.png]
{title=Ten two-element <subsets> of a five-element <set> match the corresponding entry of <Pascal's triangle>}
For a uniformly chosen two-person team, a specified person belongs to four of the ten teams, so the <probability> of their inclusion is $4/10=2/5$. Uniform selection of a $k$-element <subset> similarly gives inclusion <probability> $k/n$ for each member when $n>0$, as the <double counting> identity shows.
Back to article page
