Pascal's Rule

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Theorem

Let $\displaystyle \binom n k$ be a binomial coefficient.


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

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

This is also valid for the real number definition:

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


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

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


Direct Proof

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

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

$\blacksquare$


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

From Factors of Binomial Coefficients:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle \left({r + 1}\right) \binom r {k - 1} + \left({r + 1}\right) \binom r k\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle k \binom {r + 1} k + \left({r + 1 - k}\right) \binom {r + 1} k\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \left({r + 1}\right) \binom {r + 1} k\) \(\displaystyle \) \(\displaystyle \)                    

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

$\blacksquare$


Also see


Source of Name

This entry was named for Blaise Pascal.


Sources