0

I want to solve the following exercise:

Give a combinatorial proof of the identity $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$. Hint: Think about the meaning behind $\binom{n}{k}$ and try to separate it into two parts.

Approach: Given a set $S$ with $|S|=n$, $\binom{n}{k}$ is the number of k-elemental subsets of $S$. For simplicity let $S=\{1,...,n\}$.

I did a short example for $S=\{1,2,3\}$: $$ 2^S = \bigl\{ \emptyset, \{1\}, \{2\}, \{3\}, \{1,2\}, \{2,3\}, \{1,3\}, \{1,2,3\} \bigr\}. $$

Based on that I thought that we can consider the collection $C_1$ of subsets $2$-elemental subsets of $S$ that do not contain the $n$-th element, i.e., $C_1 = \bigl\{ \{1,2\} \bigr\}$. Thus $C_1$ contains the $2$-elemental subsets of $2^{S \setminus 3}$. (Well, maybe the example I chose was a little bit too simple.)

Now $|C_1|$ corresponds to $\binom{n-1}{k}$.

Next we can consider the 2-elemental subsets that contain $3$, $C_2 := \bigl\{ \{2,3\}, \{1,3\} \bigr\}$.

If I ignore the $3$, I get $C'_2 := \bigl\{ \{2\}, \{1\} \bigr\}$, which corresponds to the $1$-elemental subsets of $2^{S \setminus 3}$, thus $\binom{n-1}{k-1}$.

Since $|C_1 \cup C_2| = \binom{n}{k}$ and $C_1,C_2$ are disjoint, we get $$ \binom{n}{k} = |C_1 \cup C_2| = |C_1 \cup C'_2| = |C_1| \cup |C'_2| = \binom{n-1}{k} + \binom{n-1}{k-1}. $$

I have some questions: Obviously the first one is, if the solution is correct. The second one is: In the second half, I only thought of ignoring $3$ in the elements of $C_2$ because I knew what identity I had to show. If the question would just have been "find a recursive binomial formula for $\binom{n}{k}$", I would have not thought of this "trick". For that reason, are there any other useful tricks regarding combinatorial proofs? (I know this might be a too open ended question.)

Sammy Black
  • 28,409
NTc5
  • 1,161
  • Not following. You seem to be working a particular example, not the general case. And your proof for the special case looks purely enumerative, not combinatorial. – lulu Sep 24 '24 at 16:24
  • 1
    Yes, you have the right idea insofar as things go in the $n = 3$ case; the trick is to generalize it for arbitrary $n$ and $k$. In general, the key for these kinds of proofs is a recursive characterization. In other words, since we're talking about $k$-element subsets of ${1,\dots,n}$, the natural question that leads to this proof is "how can I use a subset of ${1,\dots,n-1}$ to build a subset of ${1,\dots,n}$?". – Ben Grossmann Sep 24 '24 at 16:24
  • 2
    In any case, this question has been asked repeatedly on the site. Here for instance. – lulu Sep 24 '24 at 16:26
  • Note that for each choice of $k$ elements out of $n$ elements, you can choose the last element say, and also $k-1$ out of the $n-1$ elements left, or you can skip the last one, and choose the $k$ elements out of the $n-1$ left. This gives you the identity. – Omer Simhi Sep 24 '24 at 16:27
  • Nitpicks: in your final chain of equations, (1) it's not that $C_2 = C'_2$, but rather that $|C_2| = |C'_2|$, and (2) cardinality of sets are natural numbers so we add them (not union); so it should read $$ |C_1 \cup C_2| = |C_1| + |C_2| = |C_1| + |C'_2|. $$ – Sammy Black Sep 24 '24 at 16:30
  • Also, have a look here or here or here. – Sammy Black Sep 24 '24 at 16:37

0 Answers0