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I've recently seen a certain proof which used Binomial theorem to certify the following equation. Assume $F=G\cup H$.

$$\sum_{I\subseteq G}(1-x)^{|I|}x^{|F|-|I|} = x^{|H|}$$

In the original text it was $e^{-a}$ instead of $x$ but I assumed it generalizes to the above form.

Is this equation valid? I can't really see why it holds, could someone point me towards the right direction?

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    Where is the $I$ in the addend? – 2017-01-07
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    The sum is over $I$, but $I$ does not appear in the summand. – 2017-01-07
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    Oh, oh, right, right... A typo – 2017-01-07

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Let $n=|G|$. For $0\le k\le n$ there are $\binom{n}k$ subsets $I$ of $G$ such that $|I|=k$, so

$$\begin{align*} \sum_{I\subseteq G}(1-x)^{|I|}x^{|F|-|I|}&=\sum_{k=0}^n\binom{n}k(1-x)^kx^{|F|-k}\\ &=x^{|H|}\sum_{k=0}^n\binom{n}k(1-x)^kx^{|G|-k}\\ &=x^{|H|}\sum_{k=0}^n\binom{n}k(1-x)^kx^{n-k}\\ &=x^{|H|}\cdot1^n\\ &=x^{|H|}\;, \end{align*}$$

assuming that $G\cap H=\varnothing$.