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. So far (and on our warm-up worksheet problem), we've developed our skills at counting the number of outcomes in certain scenarios. In this lecture, we'll do the reverse, translating numerical expressions into scenarios to show that two expressions are valid ways to count the same set.
For example, consider the following theorem:
For all integers $n\geq 0$ and $0 \leq k \leq n$, $$ \binom{n}{k} = \binom{n}{n-k}. $$
We will show that each side of this expression counts the number of ways to select $k$ people from a group of $n$ people. The left-hand side counts the number of ways to select $k$ people from $n$ people directly, while the right-hand side counts this by the following decision process:
For another 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 $$ {n \choose 0} + {n \choose 1} + {n \choose 2} + {n \choose 3} = 2^3. $$ By our definition of binomial coefficients, $$ \frac{3!}{3!\cdot 0!} + \frac{3!}{2!\cdot 1!} + \frac{3!}{1!\cdot 2!} + \frac{3!}{0!\cdot 3!} = 2^3. $$
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 subsets of any set of size $n$. Let's say our set of size $n$ is called $X$. Now consider the left-hand side. For each $0\leq i \leq n$, $\binom{n}{i}$ is the number of subsets of size $i$ in $X$. Every subset $A \subseteq X$ has size $0\leq |A| \leq n$, so the left-hand side is counting these sets as well. Both sides of the equation count the number of subsets of a set of size $n$, so they are equal.
The right-hand side counts the number of binary strings of length $n$. Now consider the left-hand side. For each $0\leq i \leq n$, the number of binary strings of length $n$ with $i$ 1's is $\binom{n}{i}$ — first we place the 1's ($\binom{n}{i}$ choices), then place the rest of the 0's (1 choice). Every binary string of legnth $n$ has between 0 and $n$ 1's, so the left-hand side is separately counting the number of binary strings in each of these cases. Both sides of the equation count the number of binary strings of length $n$, so they are equal.
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 $n, k \gt 0$, $$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}.$$
Let $A$ be a set of $n$ elements, where $n > 0$. The left-hand side counts 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 choose 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 $a$. Therefore both sides of the equation count the number of subsets of size $k$ from a set of size $n$, so they are equal.
This identity leads us to the famous Pascal's Triangle, a visual way to organize the binomial coefficients. 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).
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 $k$.
The right-hand side also counts the number of subsets of $A \cup B$ of size $k$ by the following decision process:
Observe that any subset of $k$ elements from $A\cup B$ has $i$ elements from $A$ and $k-i$ elements from $B$ for some $i$, so this process counts every outcome exactly once.
Therefore both sides of the equation count the number of subsets of size $k$ from $A \cup B$, so they are equal.
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.