# Pascal's Rule/Combinatorial Proof

## Theorem

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:

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

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

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

Hence the result.

$\blacksquare$

## Source of Name

This entry was named for Blaise Pascal.

## Sources

- 1925: George Pólya and Gábor Szegő:
*Problems and Theorems in Analysis I*: $\S 1.2$: Problem $31.1$ - 1966: Richard A. Dean:
*Elements of Abstract Algebra*... (previous) ... (next): $\S 0.6$: Theorem $8: \ 3$ - 1971: George E. Andrews:
*Number Theory*... (previous) ... (next): $\text {3-1}$ Permutations and Combinations: Exercise $3$