Here's how it is. So I was studying some notes on Discrete Structures then I discovered this. Prove that $2^N = \binom{N}{0} + \binom{N}{1} + \binom{N}{2} + \dots + \binom{N}{N}$ Since I was new to this, I didn't quite understand it. I tried using the binomial expansion but it confused me a little. Can someone please help me solve this. That would be great.
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Assuming that $N\in \mathbb{N}$, and by $(N,x)$ you meant $\binom{N}{x}$ then
$$ 2^N = (1+1)^N = \sum_{x=0}^N\binom{N}{x}1^x1^{N-x} = \sum_{x=0}^N\binom{N}{x} $$
V. Vancak
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I am giving a combinatorial proof.
Consider a set $S$ with $n$ elements, i.e. #$\color{blue}{S=n}$.
Now, try to count the number subsets we can form.
It, is easily found that number of $1$-subsets is $\color{blue}{n\choose 1}$, number of $2$-subsets is $\color{blue}{n\choose 2}$, $\dots$, number of $n$-subsets is $\color{blue}{n\choose n}$, and the number of null subset is $\color{blue}{n\choose 0}$ .
So, the total number of subsets is $$\color{blue}{{n\choose 0}+{n\choose 1}+{n\choose 2}+\dots+{n\choose n}}.$$
Now, try to count another way. Suppose that we have a basket containing subsets of $S$. Now pick up each elements of $S$ and we have $2$ choices to place them, in the basket of back to the set $S$. So, for every element, there are $2$ choices, so, number of ways is $\color{blue}{2^n}$.
Now, since we counted the same thing in two different ways, so the different expressions we got must be equal.
So,$$\color{red}{{n\choose 0}+{n\choose 1}+{n\choose 2}+\dots+{n\choose n}=2^n}.$$
(N, 0), do you mean $N \choose 0$? Also, does it stop at $N \choose 11$ or does it go all the way up to $N \choose N$? – Noble Mushtak Jan 24 '16 at 14:23\chooseto\binom.\choose(as all TeX-style fraction commands) are obsolete and should not be used anymore. Just for the record.{N\choose0}->\binom{N}{0}. And don't worry for missing that :). – MickG Jan 24 '16 at 15:01