Sum of Binomial distribution

Question

I am struggling to show the following -

$$\sum_{N_{H}=0}^{N} \frac{N!}{N_{H}!(N-N_{H}!)}f_{H}(1-f_{H})^{N-N_{H}} = (f_{H}+(1-f_{H}))^N$$

I would like to prove it without doing induction. I'm sure it's quite simple but I can't seem to figure it out!

Thanks for the help.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{N_{H} = 0}^{N}{N! \over N_{H}!\pars{N - N_{H}!}} \,f_{H}\pars{1 - f_{H}}^{N - N_{H}} & = f_{H}\sum_{N_{H} = 0}^{N}{N \choose N_{H}}\,\pars{1 - f_{H}}^{N_{H}} = f_{H}\bracks{1 + \pars{1 - f_{H}}}^{N} \\[5mm] & = \bbx{\ds{f_{H}\pars{2 - f_{H}}^{N}}} \end{align}