What is combinatorial argument

We answered the same question in two different ways, so the two answers must be the same. You can also explain prove this identity by counting subsets, or even lattice paths. Why is this true? Are these really the same? So the set of outcomes should be the same. Let's try the pizza counting example like we did above. Alternatively, you could make a list of all the toppings you don't want. You can also prove explain this identity using bit strings, subsets, or lattice paths.

On the other hand, divide the possible pizzas into disjoint groups: the pizzas with no toppings, the pizzas with one topping, the pizzas with two toppings, etc. How many pizzas have 1 topping? We have:. The total number of possible pizzas will be the sum of these, which is exactly the left-hand side of the identity we are trying to prove.

Again, we could have proved the identity using subsets, bit strings, or lattice paths although the lattice path argument is a little tricky. Hopefully this gives some idea of how explanatory proofs of binomial identities can go.

It is worth pointing out that more traditional proofs can also be beautiful. W e will discuss indu ction in Section 2.

For example, consider the following rather slick proof of the last identity. The explanatory proofs given in the above examples are typically called combinatorial proofs. The tricky thing is coming up with the question. This is not always obvious, but it gets easier the more counting problems you solve. You will start to recognize types of answers as the answers to types of questions.

More often what will happen is you will be solving a counting problem and happen to think up two different ways of finding the answer. Now you have a binomial identity and the proof is right there. We just showed that the left-hand side and the right-hand side of the given identity are two different ways of counting the committee-chair possibilities, and hence, they must be equal. While the combinatorial proof of the Chairperson Identity is no more correct than the algebraic method, it offers a concrete, meaningful way to explain why the two quantities are always equal.

Proving combinatorial identities in this manner requires creativity, especially if one is not told what set is being counted. Prove the statement in Checkpoint 2. Section 2. Learn more. How to prove it by means of a combinatorial argument? A combinatorial exercise [duplicate] Ask Question. Asked 9 years, 7 months ago. Active 9 years, 7 months ago. Viewed 10k times. Community Bot 1. Add a comment. What if the tournament goes all 7 games? So you win the last game.

How many ways can the first 6 games go down? What if the tournament goes just 6 games? How many ways can this happen? What about 5 games? What are the two different ways to compute the number of ways your team can win? What pattern in Pascal's triangle is this an example of? Have a look again at Pascal's triangle. Forget for a moment where it comes from.

Just look at it as a mathematical object. What do you notice? There are lots of patterns hidden away in the triangle, enough to fill a reasonably sized book. Here are just a few of the most obvious ones:. The triangle is symmetric. In any row, entries on the left side are mirrored on the right side. We would like to state these observations in a more precise way, and then prove that they are correct. Now each entry in Pascal's triangle is in fact a binomial coefficient.

Given this description of the elements in Pascal's triangle, we can rewrite the above observations as follows:. Each of these is an example of a binomial identity : an identity i. Our goal is to establish these identities. This is certainly a valid proof, but also is entirely useless.

Even if you understand the proof perfectly, it does not tell you why the identity is true. Let's see how this works for the four identities we observed above. What do these binomial coefficients tell us? There is only one way to do this, namely to not select any of the objects.

There is only one such string, the string of all 0's. There is only one such subset, the empty set. The only such subset is the original set of all elements. The easiest way to see this is to consider bit strings. Of all of these strings, some start with a 1 and the rest start with a 0.