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I saw this question: The number of odd size subsets is equal to the number of even size subsets but I'm trying to prove it at a different way:
I'm trying to prove that: $$\sum_{i=0}^{\left\lceil \frac{n}{2}\right\rceil }{n \choose 2i}=\sum_{i=0}^{\left\lceil \frac{n}{2}\right\rceil }{n \choose 2i+1}$$

if $n$ is odd, I can do it with the identity: ${n \choose k}={n \choose n-k}$, because if we have odd number like $7$ we have $4$ ($=\left\lceil \frac{n}{2}\right\rceil$) pairs: $(7,0),(6,1),(5,2),(4,3)$. The left element is odd and the right one is even. So:
$${7 \choose 0}={7 \choose 7},\,{7 \choose 6}={7 \choose 1},...$$

But when I'm trying to do this about even numbers (like $8$), I can't use this method.
So my question is: What can I do at cases of even numbers?

Thank you!

Moo
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CS1
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2 Answers2

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The claim is only true for non-empty sets $A$. There's a reason why the observation that (for any fixed element $a\in A$) $$ X\mapsto X\operatorname\Delta\{a\}$$ is a bijection from the even-sized subsets of $A$ to the odd-sized subsets (as well as vice versa) is considered a simpler proof of the desired claim.

Hagen von Eitzen
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  • Thank you, but does it means? $X\mapsto X\operatorname\Delta{a}$ and whats is $X$? – CS1 Aug 07 '21 at 11:43
  • $X$ is an arbitrary subset of $A$. And $\Delta$ is the symmetric difference. It means you add ${a}$ to $X$, if $a\notin A$ and exclude it otherwise, i.e. $f(X)=\begin{cases} X\cup{a}, &a\notin X\ X\setminus{a},&a\in X\end{cases}$. Hence, $f$ maps the odd subsets bijectively to the even ones. – Jochen Aug 07 '21 at 12:00
  • A bit more verbose explanation: pick any element $a$ from the given set and keep it fixed. Now, if you have a subset $X$, transform it: either add $a$ (if it wasn't there) or take $a$ out (if it was). That turns an odd-sized set $X$ into an even-sized set and vice versa. –  Aug 07 '21 at 12:37
  • Now I understand it better, but how it's help me to prove it? @StinkingBishop - it's for both even and odd sets, right? – CS1 Aug 07 '21 at 12:53
  • @Jochen - Can you explain me more please how does it helps to prove it? – CS1 Aug 07 '21 at 13:39
  • @CS1 If there is a bijection between two collections then you know they are of the same size. Since there is a bijection from the collection of even-sized subsets to the collection of odd-sized subsets, we know they must be of the same size. It follows then that half of all subsets of an $n$-element set ($n\geq 1$) are even-sized and half are odd-sized and so since there are $2^n$ subsets of an $n$-element set it follows that $2^{n-1}$ are even-sized. – JMoravitz Aug 07 '21 at 13:41
  • @CS1: The odd-sized sets with $a$ in them are paired up with even-sized sets without $a$, and, similarly, the even-sized sets with $a$ are paired up with odd-sized sets without $a$. Does that explanation make it any clearer? –  Aug 08 '21 at 07:57
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I shall show how to prove this in the case $n=8$, and it should be clear how this idea generalizes. You need to prove $$ \binom80+\binom82+\binom84+\binom86+\binom88\stackrel{?}=\binom81+\binom83+\binom85+\binom87 $$ First, apply Pascals rule to each entry in the above purported equation except for $\binom80$ and $\binom88$, which says that $\binom{8}{k}=\binom{7}k+\binom7{k-1}$. The result is $$ \begin{align} \binom80+\left[\binom71+\binom72\right]+\left[\binom73+\binom74\right]+\left[\binom75+\binom76\right]+\binom88=\\ \left[\binom70+\binom71\right]+\left[\binom72+\binom73\right]+\left[\binom74+\binom75\right]+\left[\binom76+\binom77\right] \end{align} $$ After realizing that $\binom80=\binom70$ and $\binom88=\binom77$, the last equation should be obvious.

Mike Earnest
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  • And... in fact this is the same proof as the other one. E.g ${8\choose 2}={7\choose 1}+{7\choose 2}$ can be interpreted as breaking all 2-element subsets of the 8-element set into those ${7\choose 1}$ which contain a fixed element $a$ and those ${7\choose 2}$ subsets which don't contain it! –  Aug 08 '21 at 08:07