Question Number 207387 by sniper237 last updated on 13/May/24
$${Let}\:\:{cardE}={n}\:,\:{and}\:\:{the}\:{set}\:{of}\:{parts} \\ $$$${S}=\left\{\left({A},{B}\right)\in{P}\left({E}\right)×{P}\left({E}\right)\:/\:\:{A}\cap{B}=\varnothing\right\} \\ $$$${Show}\:{that}\:\:{cardS}=\:\mathrm{3}^{{n}} \\ $$
Answered by Berbere last updated on 13/May/24
$${if}\:{card}\left({A}\right)={k};{E}={A}\cup\overset{−} {{A}}\:\:{the}\:{number}\:{of}\:{subset}\:{of}\:{card}={k} \\ $$$${in}\:{E}\:{is}\:\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix} \\ $$$${we}\:{havd}\:{to}\:{shoose}\:{B}\:{in}\:\left(\overset{−} {{A}}\right);{card}\left(\overset{−} {{A}}\right)={n}−{k} \\ $$$${B}\in{P}\left(\overset{−} {{A}}\right)\:{card}\:\left({P}\left(\overset{−} {{A}}\right)\right)=\mathrm{2}^{{n}−{k}} \\ $$$$\left({A},{B}\right)\:{can}\:{bee}\:{chosed}\:{by}\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\mathrm{2}^{{n}−{k}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\mathrm{1}^{{k}} .\mathrm{2}^{{n}−{k}} =\left(\mathrm{1}+\mathrm{2}\right)^{{n}} =\mathrm{3}^{{n}} \\ $$$$ \\ $$$$ \\ $$