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I am trying to prove that

$\sum_{k=0}^i\binom{i}{k} = 2^i$ by counting in two ways.

i.e. count the subsets of an $i-set$ in two ways.

So far I have that the total number of subsets of an $i-set$ is $\sum_{k=0}^{i}\binom{i}{k}$

now for the second part let $R=\{a_1,a_2,...,a_i\}$ be an $i-set$ and let $A_k \subset R $

so for $a_j, j=1,2,...,i$ either $a_j \in A_k$ or $a_j \notin A_k$

I'm trying to use this to show that this give two choices for $a_j$ in $A_k$ and then use this to show that choosing each element in $A_k \forall \space k = 1,2,...,i$ is equivalent to $2^i$ but my argument is not so clear. Any help would be much appreciated.

Pablo
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2 Answers2

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As you have very well demonstrated the number of ways of choosing a subset from $n$ elements is $$\sum_{i=0}^n \binom{n}{i}$$

Now there is another strategy we can adopt to choose subsets, for every element in our original set, we have two options, either include it in our new subset, or discard it. This means we have $2^n$ options in total, justified by The fundamental principle of counting.

Therefore

$$\sum_{i=0}^n \binom{n}{i}=2^n$$

Although personally I prefer substituting $x=1$ in the expansion for $(1+x)^n$

Guy
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  • using the binomial expansion would be so much easier, but I had to prove this counting in two ways so I couldn't use it. – Pablo Mar 18 '14 at 21:00
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    @Henry double counting leads to easy and wonderful proofs sometimes with deep insights as to why two unrelated things are equal. get habituated, it will come in very handy. – Guy Mar 18 '14 at 21:02
  • it is quite elegant I must say, it's the first time I'm doing a combinatorics course and the combinatorial approach to proof is very eloquent if a little difficult to get used to. thanks again for your help Sabyasachi – Pablo Mar 18 '14 at 21:04
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    @Henry for instance, if I shamelessly brag, see this – Guy Mar 18 '14 at 21:05
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    One nice insight you will get from the binomial theorem proof is into thinking more deeply about ordinary generating functions. One (perhaps unrelated) thing to consider with the binomial theorem, is what would happen if you tried to use that argument and $\mathbb{Z}$ wasn't Abelian (commutative)? – ml0105 Mar 18 '14 at 21:05
  • @ml0105 ah. very true. math is wonderful isn't it? :D – Guy Mar 18 '14 at 21:07
  • Quite beautiful indeed! I love seeing algebra come into play with combinatorics and graph theory, as well as looking at algebra with a combinatorial mindset. There is usually a lot of deep stuff going on under the hood. – ml0105 Mar 18 '14 at 21:08
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    @Sabyasachi I say sir, you may shamelessly brag:D – Pablo Mar 18 '14 at 21:10
  • @ml0105 although I know something is true when I write it, and therefore no matter how I get to the result, I must always arrive at the same conclusion, it still amazes me when different processes give the same answer. – Guy Mar 18 '14 at 21:10
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I would look at this with the binary string argument. Consider vectors in $\{0, 1}^{i}$. If the $jth$ position is $1$, then $a_{j}$ is in the subset. Otherwise, $a_{j}$ is not in the subset. As each position is independent of the others and we have two options for each position, we have $2^{i}$ possible subsets of $A$.

Does that help clarify?

ml0105
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