# Pascal's rule

In mathematics, Pascal's rule is a combinatorial identity about binomial coefficients. It states that for any natural number n we have

${\displaystyle {n-1 \choose k}+{n-1 \choose k-1}={n \choose k}\quad {\text{for }}1\leq k\leq n}$

where ${\displaystyle {n \choose k}}$ is a binomial coefficient. This is also commonly written

${\displaystyle {n \choose k}+{n \choose k-1}={n+1 \choose k}\quad {\text{for }}1\leq k\leq n+1}$

## Combinatorial proof

Pascal's rule has an intuitive combinatorial meaning. Recall that ${\displaystyle {a \choose b}}$ counts in how many ways can we pick a subset with b elements out from a set with a elements. Therefore, the right side of the identity ${\displaystyle {n \choose k}}$ is counting how many ways can we get a k-subset out from a set with n elements.

Now, suppose you distinguish a particular element 'X' from the set with n elements. Thus, every time you choose k elements to form a subset there are two possibilities: X belongs to the chosen subset or not.

If X is in the subset, you only really need to choose k − 1 more objects (since it is known that X will be in the subset) out from the remaining n − 1 objects. This can be accomplished in ${\displaystyle n-1 \choose k-1}$ ways.

When X is not in the subset, you need to choose all the k elements in the subset from the n − 1 objects that are not X. This can be done in ${\displaystyle n-1 \choose k}$ ways.

We conclude that the numbers of ways to get a k-subset from the n-set, which we know is ${\displaystyle {n \choose k}}$, is also the number ${\displaystyle {n-1 \choose k-1}+{n-1 \choose k}.}$

## Algebraic proof

We need to show

${\displaystyle {n \choose k}+{n \choose k-1}={n+1 \choose k}.}$

{\displaystyle {\begin{aligned}{n \choose k}+{n \choose k-1}&={\frac {n!}{k!(n-k)!}}+{\frac {n!}{(k-1)!(n-k+1)!}}\\&=n!\left[{\frac {n-k+1}{k!(n-k+1)!}}+{\frac {k}{k!(n-k+1)!}}\right]\\&={\frac {n!(n+1)}{k!(n-k+1)!}}={\binom {n+1}{k}}\end{aligned}}}

## Generalization

Let ${\displaystyle n,k_{1},k_{2},k_{3},\dots ,k_{p},p\in \mathbb {N} ^{*}}$ and ${\displaystyle n=k_{1}+k_{2}+k_{3}+\cdots +k_{p}}$. Then

{\displaystyle {\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}}}