Pascal's Rule

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

For positive integers $$n, k \,\!$$ with $$1 \le k \le n \,\!$$:
 * $$\binom n {k-1} + \binom n k = \binom {n+1} k$$

This is also valid for the real number definition:


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

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 $$\binom {n+1} k$$ 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 $$\binom n k$$ 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 $$\binom n {k-1}$$ ways of selecting the other $$k-1$$ members so as to form such a committee.

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