## Online Course Discussion Forum

### Math Challenge II-A Combinatorics 6.30

Math Challenge II-A Combinatorics 6.30

Math Challenge II-A Combinatorics 6.30:

How is the midterm determined? And how is symmetry used? I am really stuck on this chapter. It is too hard.

Thanks

Re: Math Challenge II-A Combinatorics 6.30

One thing that can definitely help is looking at some examples for these more abstract questions.

Let's take a look here at $n = 2$. Thus, $4n+2 = 10$ and $2n = 4$. We want to show: $$\binom{10}{0} - \binom{10}{1} + \binom{10}{2} - \binom{10}{3} + \binom{10}{4} = \frac{1}{2} \binom{10}{5}.$$ Following, the hint, let's start by considering $(1+x)^{10}$ with $x = -1$. In fact, this is a result that was done before in the examples, and gives us that $$\binom{10}{0} - \binom{10}{1} + \binom{10}{2} - \binom{10}{3} \cdots + \binom{10}{8} - \binom{10}{9} + \binom{10}{10} = 0.$$ Remember by symmetry $\displaystyle \binom{10}{k} = \binom{10}{10-k}$, so we can pair up terms: $$\binom{10}{0} = \binom{10}{10}, \binom{10}{1} = \binom{10}{9}, \text{ etc.}$$ Note that all the terms pair up EXCEPT when $k=5$. Therefore, the above expression can be rewritten as $$2\binom{10}{0} - 2\binom{10}{1} + 2\binom{10}{2} - 2\binom{10}{3} +2\binom{10}{4} - \binom{10}{5} = 0.$$ Moving the $\displaystyle \binom{10}{5}$ to the other side and dividing by $2$ gives us the result we need.

In fact, the general proof (when $n$ is a variable) is exactly the same argument, you just need to be careful with making sure you understand the pattern. Try writing it out for $n=3$, and if you can understand that, you probably understand it in general!

Thanks!