Combinatorial proofs are a nice form of proof that embody the core of what we've been doing this unit: counting and analyzing objects/sets/events by relating them to other objects/sets/events. In this lesson, rather than taking an event/set that we can't count and relating it to an event that we know how to count, we'll take a slightly bigger step: we'll show that two mathematical expressions are equivalent by showing that they're both valid ways to count the same set!
For example, consider the following theorem.
For all positive integers $n$, $$\sum_{i=0}^n \binom n i = 2^n.$$
In some sense, this is a very complex statement; One is summing up binomial coefficients, which individually have a complicated formula. For $n=3$, it is asserting that $$ \frac{3!}{3!\cdot 0!} + \frac{3!}{2!\cdot 1!} + \frac{3!}{1!\cdot 2!} + \frac{3!}{0!\cdot 3!} = 2^3. $$
The fact that terms with lots of factorials always collapses is remarkable. On the other hand, the formula has a very simple explanation: The right-hand side is counting the number of subsets of a set of size $n$. It turns out that the left-hand is also counting the same sets! It's counting the $\binom{n}{0}$ sets of size zero, the $\binom{n}{1}$ sets of size one, and so on up to the maximum size.
This type of reasoning, where one avoids algebra and/or induction and just proves an identity via counting, is called a combinatorial proof (The BH textbook calls these "story proofs.) They are completely rigorous and acceptable, and often preferred to algebraic proofs because they usually come with a built-in explanation for why the identity is true. Here is an example of how to write up the above reasoning:
The right-hand side counts the number of sets of any set of size $n$ (for concreteness, let's fix a set $X$ of size $n$). Now consider the left-hand size. For each $0\leq i \leq n$, $\binom{n}{i}$ is the number of subsets of size $i$ in $X$. But every subset $A \subset X$ has size $0\leq |A| \leq n$, so the left-hand side is counting these sets as well.
There are other proofs, say by induction. We'll do a different proof below using something called the Binomial Theorem.
Here is another simple example of a combinatorial proof.
For all integers $n\geq 0$ and $0 \leq k \leq n$, $$ \binom{n}{k} = \binom{n}{n-k}. $$
You can prove this instantly by algebra. A combinatorial proof just observes that the left-hand side counts the number of subsets of size $k$. The right-hand is doing the same thing, following this decision process:
This establishes the theorem.
There are a lot of other neat results that are provable with combinatorial proofs. Perhaps the most famous one is due to Blaise Pascal in the 1600s.
For all integers numbers $n \gt k \gt 0$, $$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}.$$
There is a straightforward algebraic proof of this fact, but let's do a combinatorial proof.
Let $A$ be a set of $n \geq 2$ elements. Left-hand side count the number of subsets of $A$ of size $k$, which is $\binom{n}{k}$. We must show that right-hand side also counts these sets.
Since $A$ is non-empty, we can fix an arbitrary element of $A$, say $a$. Let $r$ be the number of subsets of $A$ that contain $a$, and $s$ be the number of subsets that don't. Then $$ \binom{n}{k} = r + s $$ because every subset either contains $a$ or doesn't.
We claim that $r=\binom{n-1}{k-1}$. Here is a decision process for counting sets that contain $a$:
We next claim that $s=\binom{n-1}{k}$. This has a one-step decision process: Just pick $k$ elements from the $n-1$ elements of $A$ that are not $a$. This completes the proof.
This identity leads us to the famous Pascal's Triangle, where we start by arranging the nontrivial binomial coefficients in the following infinite pattern: $$ \begin{matrix} &&&&&& \binom 0 0 &&&&&& \\ &&&&& \binom 1 0 && \binom 1 1 &&&&& \\ &&&& \binom 2 0 && \binom 2 1 && \binom 2 2 &&&& \\ &&& \binom 3 0 && \binom 3 1 && \binom 3 2 && \binom 3 3 &&& \\ && \binom 4 0 && \binom 4 1 && \binom 4 2 && \binom 4 3 && \binom 4 4 && \\ & \binom 5 0 && \binom 5 1 && \binom 5 2 && \binom 5 3 && \binom 5 4 && \binom 5 5 & \\ \binom 6 0 && \binom 6 1 && \binom 6 2 && \binom 6 3 && \binom 6 4 && \binom 6 5 && \binom 6 6 \end{matrix} $$ Filling in the values for the coefficients, the triangle looks like this: $$ \begin{matrix} &&&&&& 1 &&&&&& \\ &&&&& 1 && 1 &&&&& \\ &&&& 1 && 2 && 1 &&&& \\ &&& 1 && 3 && 3 && 1 &&& \\ && 1 && 4 && 6 && 4 && 1 && \\ & 1 && 5 && 10 && 10 && 5 && 1 & \\ 1 && 6 && 15 && 20 && 15 && 6 && 1 \end{matrix} $$ When viewed like this, it's easy to see how Pascal's identity is used to construct the following row: We just add the two adjacent entries and put the sum in the next row between them (and put $1$'s on the ends).
This arrangement leads us to even more interesting identities or interpretations of identities with respect to the triangle. For example, Theorem 15.1 is just the sum of the $n$th row of the triangle.
The following identity is due to Alexandre-Théophile Vandermonde from the late 1700s.
For all $n,m \geq k \gt 0$, $$ \binom{m+n}{k} = \sum_{i=0}^k \binom{m}{i} \binom{n}{k-i}.$$
If we have disjoint sets $A$ and $B$ with $|A| = m$ and $|B| = n$, then $\binom{m+n}{k}$ is the number of subsets of $A \cup B$ of size $k$.
We can consider this decision process choosing an $k$-element subset of $A \cup B$. Beware that this decision process gives a tree that is not "regular", meaning that we cannot directly use the Multiplication Rule.
To see that this process counts every outcome exactly once, observe that any subset of $k$ elements from $A\cup B$ has $i$ elements from $A$ and $k-i$ elements from $B$ for some $i$.
Note that our decision in the first stage affects the number of options in the later stages, so we need to be careful. After we select $i$ however, the rest is regular: There are $\binom{m}{i}\binom{n}{k-i}$ leaves under this part of our tree. Now we apply the Addition Rule, and add up these counts as $i$ ranges from $0$ to $k$ to get the theorem.
The following interesting fact follows from Vandermonde's identity.
Using Vandermonde's identity, we set $m = k = n$ to get $$\binom{2n}{n} = \sum_{i=0}^n \binom{n}{n-k} \binom n k.$$ Then we observe that $\binom{n}{n-k} = \binom n k$.