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

  • 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text I$. Algebra: Permutations and Combinations: Two important relations
  • 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 $3$
  • 1972: George Pólya and Gábor Szegő: Problems and Theorems in Analysis I (Eng. ed.): $\S 1.2$: Problem $31.1$
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