# Pascal's Rule

## Theorem

Let $\dbinom n k$ be a binomial coefficient.

For positive integers $n, k$ with $1 \le k \le n$:

$\dbinom n {k-1} + \dbinom n k = \dbinom {n+1} k$

This is also valid for the real number definition:

$\forall r \in \R, k \in \Z: \dbinom r {k-1} + \dbinom r k = \dbinom {r+1} k$

Thus the binomial coefficients can be defined using the following recurrence relation:

$\dbinom n k = \begin{cases} 1 & : k = 0 \\ 0 & : k > n \\ \dbinom{n-1}{k-1} + \dbinom{n-1}{k} & : \text{otherwise} \end{cases}$

## Direct Proof

Let $n, k \in \N$ with $1 \le k \le n$.

 $$\displaystyle \binom n k + \binom n {k-1}$$ $$=$$ $$\displaystyle \frac{n!}{k! \ \left({n - k}\right)!} + \frac{n!}{\left({k - 1}\right)! \ \left({n - \left({k - 1}\right)}\right)!}$$ Definition of Binomial Coefficient $$\displaystyle$$ $$=$$ $$\displaystyle \frac{n! \ \left({n - \left({k - 1}\right)}\right)} {k! \ \left({n - k}\right)! \ \left({n - \left({k - 1}\right)}\right)} + \frac{n! \ k}{\left({k - 1}\right)! \ \left({n - \left({k - 1}\right)}\right)! \ k}$$ $$\displaystyle$$ $$=$$ $$\displaystyle \frac{n! \ \left({n - k + 1}\right)} {k! \ \left({n - k + 1}\right)!} + \frac {n! \ k}{k! \ \left({n - k + 1}\right)!}$$ $$\displaystyle$$ $$=$$ $$\displaystyle \frac{n! \ \left({n - k + 1}\right) + n! \ k} {k! \ \left({n - k + 1}\right)!}$$ $$\displaystyle$$ $$=$$ $$\displaystyle \frac{n! \ \left({n - k + 1 + k}\right)} {k! \ \left({n - k + 1}\right)!}$$ $$\displaystyle$$ $$=$$ $$\displaystyle \frac{n! \ \left({n + 1}\right)}{k! \ \left({n - k + 1}\right)!}$$ $$\displaystyle$$ $$=$$ $$\displaystyle \frac{\left({n + 1}\right)!} {k! \ \left({n + 1 - k}\right)!}$$ $$\displaystyle$$ $$=$$ $$\displaystyle \binom {n+1} k$$

$\blacksquare$

## Combinatorial Proof

Suppose you were a member of a club with $n + 1$ members (including you).

Suppose it were time to elect a committee of $k$ members from that club.

From Cardinality of Set of Subsets, there are $\displaystyle \binom {n+1} k$ ways to select the members to form this committee.

Now, you yourself may or may not be elected a member of this committee.

Suppose that, after the election, you are not a member of this committee.

Then, from Cardinality of Set of Subsets, there are $\displaystyle \binom n k$ ways to select the members to form such a committee.

Now suppose you are a member of the committee. Apart from you, there are $k-1$ such members.

Again, from Cardinality of Set of Subsets, there are $\displaystyle \binom n {k-1}$ ways of selecting the other $k-1$ members so as to form such a committee.

In total, then, there are $\displaystyle \binom n k + \binom n {k-1}$ possible committees.

Hence the result.

$\blacksquare$

## Proof for Real Numbers

 $$\displaystyle \left({r + 1}\right) \binom r {k - 1} + \left({r + 1}\right) \binom r k$$ $$=$$ $$\displaystyle \left({r + 1}\right) \binom r {k - 1} + \left({r + 1}\right) \binom r {r - k}$$ Symmetry Rule for Binomial Coefficients $$\displaystyle$$ $$=$$ $$\displaystyle k \binom {r + 1} k + \left({r - k + 1}\right) \binom {r + 1} {r - k + 1}$$ Factors of Binomial Coefficients $$\displaystyle$$ $$=$$ $$\displaystyle k \binom {r + 1} k + \left({r - k + 1}\right) \binom {r + 1} k$$ Symmetry Rule for Binomial Coefficients $$\displaystyle$$ $$=$$ $$\displaystyle \left({r + 1}\right) \binom {r + 1} k$$

Dividing by $\left({r + 1}\right)$ yields the result.

$\blacksquare$

## Source of Name

This entry was named for Blaise Pascal.