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I was wondering if the following reasoning is correct to show that $[n]$ has $2^{n-1}$ odd sized subsets.

By the Binomial Theorem, $$\sum_{k=0}^n\binom{n}{k}=2^n$$ which can be interpreted as the number of even sized subsets plus the number of odd sized subsets is $2^{n}$.

Also by the Binomial Theorem, $$\sum_{k=0}^n(-1)^k\binom{n}{k}=0$$ which can be interpreted as the number of even sized subset is the same as the number of odd sized subsets.

So it follows that the number of even sized subsets and odd sized subsets is both $2^{n-1}$.

I couldn't find a solution that followed this approach so I was wondering if this was also a correct way to look at things?

Moh
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    Yes, this is a correct proof of the result. I'd elaborate that it's the binomial expansion of $2^n=(1+1)^n$ and $0=(1-1)^n$ that get you those results, but the reasoning is correct. – Robert Shore Apr 07 '23 at 23:16
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    here is a duplicate (of a duplicate) that uses the same argument. I still prefer the combinatorial argument: pair two subsets iff they differ only with the element $n$. That gives a bijection between the odd and even subsets. – lulu Apr 07 '23 at 23:17

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Intuitively, $2^{n-1} = \frac{1}{2}(2^n)$. Since half of all integers are even and half are odd, it makes sense that the cardinality of half of all possible subsets is an odd number and the cardinality of the other half of all possible subsets is an even number.

Nate
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