# Pascal's rule

In mathematics, Pascal's rule (or Pascal's formula) is a combinatorial identity about binomial coefficients. The binomial coefficients are the numbers that appear in Pascal's triangle. Pascal's rule states that for positive integers n and k, ${n-1 \choose k}+{n-1 \choose k-1}={n \choose k},$ where ${\tbinom {n}{k}}$ is the binomial coefficient, namely the coefficient of the x$^k$ term in the expansion of (1 + x)$^n$. There is no restriction on the relative sizes of n and k; in particular, the above identity remains valid when n < k since ${\tbinom {n}{k}}=0$ whenever n < k.
Together with the boundary conditions ${\tbinom {n}{0}}={\tbinom {n}{n}}=1$ for all nonnegative integers n, Pascal's rule determines that ${\binom {n}{k}}={\frac {n!}{k!(n-k)!}},$ for all integers 0 ≤ k ≤ n. In this sense, Pascal's rule is the recurrence relation that defines the binomial coefficients.
Pascal's rule can also be generalized to apply to multinomial coefficients.

## Combinatorial proof

Pascal's rule has an intuitive combinatorial meaning, that is clearly expressed in this counting proof.$^{:44}$
Proof. Recall that ${\tbinom {n}{k}}$ equals the number of subsets with k elements from a set with n elements. Suppose one particular element is uniquely labeled X in a set with n elements.
To construct a subset of k elements containing X, include X and choose k − 1 elements from the remaining n − 1 elements in the set. There are ${\tbinom {n-1}{k-1}}$ such subsets.
To construct a subset of k elements not containing X, choose k elements from the remaining n − 1 elements in the set. There are ${\tbinom {n-1}{k}}$ such subsets.
Every subset of k elements either contains X or not. The total number of subsets with k elements in a set of n elements is the sum of the number of subsets containing X and the number of subsets that do not contain X, ${\tbinom {n-1}{k-1}}+{\tbinom {n-1}{k}}$.
This equals ${\tbinom {n}{k}}$; therefore, ${\tbinom {n}{k}}={\tbinom {n-1}{k-1}}+{\tbinom {n-1}{k}}$.

## Algebraic proof

Alternatively, the algebraic derivation of the binomial case follows. ${\begin{aligned}{n-1 \choose k}+{n-1 \choose k-1}&={\frac {(n-1)!}{k!(n-1-k)!}}+{\frac {(n-1)!}{(k-1)!(n-k)!}}\\&=(n-1)!\left[{\frac {n-k}{k!(n-k)!}}+{\frac {k}{k!(n-k)!}}\right]\\&=(n-1)!{\frac {n}{k!(n-k)!}}\\&={\frac {n!}{k!(n-k)!}}\\&={\binom {n}{k}}.\end{aligned}}$
An alternative algebraic proof using the alternative definition of binomial coefficients: ${\tbinom {n}{k}}={\frac {n(n-1)\cdots (n-k+1)}{k!}}$. Indeed
${\begin{aligned}{n-1 \choose k}+{n-1 \choose k-1}&={\frac {(n-1)\cdots ((n-1)-k+1)}{k!}}+{\frac {(n-1)\cdots ((n-1)-(k-1)+1)}{(k-1)!}}\\&={\frac {(n-1)\cdots (n-k)}{k!}}+{\frac {(n-1)\cdots (n-k+1)}{(k-1)!}}\\&={\frac {(n-1)\cdots (n-k+1)}{(k-1)!}}\left[{\frac {n-k}{k}}+1\right]\\&={\frac {(n-1)\cdots (n-k+1)}{(k-1)!}}\cdot {\frac {n}{k}}\\&={\frac {n(n-1)\cdots (n-k+1)}{k!}}\\&={\binom {n}{k}}.\end{aligned}}$
Since ${\tbinom {z}{k}}={\frac {z(z-1)\cdots (z-k+1)}{k!}}$ is used as the extended definition of the binomial coefficient when z is a complex number, thus the above alternative algebraic proof shows that Pascal's rule holds more generally when n is replaced by any complex number.

## Generalization

Pascal's rule can be generalized to multinomial coefficients.$^{:144}$ For any integer p such that $p\geq 2$, $k_{1},k_{2},k_{3},\dots ,k_{p}\in \mathbb {N} ^{+}\!,$ and $n=k_{1}+k_{2}+k_{3}+\cdots +k_{p}\geq 1$, ${n-1 \choose k_{1}-1,k_{2},k_{3},\dots ,k_{p}}+{n-1 \choose k_{1},k_{2}-1,k_{3},\dots ,k_{p}}+\cdots +{n-1 \choose k_{1},k_{2},k_{3},\dots ,k_{p}-1}={n \choose k_{1},k_{2},k_{3},\dots ,k_{p}}$ where ${n \choose k_{1},k_{2},k_{3},\dots ,k_{p}}$ is the coefficient of the $x_{1}^{k_{1}}x_{2}^{k_{2}}\cdots x_{p}^{k_{p}}$ term in the expansion of $(x_{1}+x_{2}+\dots +x_{p})^{n}$.
The algebraic derivation for this general case is as follows.$^{:144}$ Let p be an integer such that $p\geq 2$, $k_{1},k_{2},k_{3},\dots ,k_{p}\in \mathbb {N} ^{+}\!,$ and $n=k_{1}+k_{2}+k_{3}+\cdots +k_{p}\geq 1$. Then ${\begin{aligned}&\quad {n-1 \choose k_{1}-1,k_{2},k_{3},\dots ,k_{p}}+{n-1 \choose k_{1},k_{2}-1,k_{3},\dots ,k_{p}}+\cdots +{n-1 \choose k_{1},k_{2},k_{3},\dots ,k_{p}-1}\\&={\frac {(n-1)!}{(k_{1}-1)!k_{2}!k_{3}!\cdots k_{p}!}}+{\frac {(n-1)!}{k_{1}!(k_{2}-1)!k_{3}!\cdots k_{p}!}}+\cdots +{\frac {(n-1)!}{k_{1}!k_{2}!k_{3}!\cdots (k_{p}-1)!}}\\&={\frac {k_{1}(n-1)!}{k_{1}!k_{2}!k_{3}!\cdots k_{p}!}}+{\frac {k_{2}(n-1)!}{k_{1}!k_{2}!k_{3}!\cdots k_{p}!}}+\cdots +{\frac {k_{p}(n-1)!}{k_{1}!k_{2}!k_{3}!\cdots k_{p}!}}={\frac {(k_{1}+k_{2}+\cdots +k_{p})(n-1)!}{k_{1}!k_{2}!k_{3}!\cdots k_{p}!}}\\&={\frac {n(n-1)!}{k_{1}!k_{2}!k_{3}!\cdots k_{p}!}}={\frac {n!}{k_{1}!k_{2}!k_{3}!\cdots k_{p}!}}={n \choose k_{1},k_{2},k_{3},\dots ,k_{p}}.\end{aligned}}$