Pascal's Rule

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:


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

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

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.

Proof for Real Numbers
Follows directly from Factors of Binomial Coefficients:

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

Also see

 * Pascal's Triangle