Question Number 205070 by dimentri last updated on 07/Mar/24
$$\:\:\:\mathrm{Given}\:\begin{cases}{\mathrm{A}\cap\mathrm{B}=\:\left\{\:\mathrm{a},\:\mathrm{b}\right\}}\\{\mathrm{A}\cap\mathrm{C}\:=\:\left\{\:\mathrm{b},\:\mathrm{c}\right\}\:}\\{\mathrm{B}\cap\mathrm{C}=\:\left\{\:\mathrm{b}\:,\mathrm{d}\:\right\}}\end{cases} \\ $$$$\:\:\:\:\mathrm{then}\:\left(\mathrm{A}\cap\mathrm{C}\right)\:+\:\left(\mathrm{A}\cap\mathrm{B}\right)\:+\:\left(\mathrm{B}\cap\mathrm{C}\right) \\ $$
Commented by JDamian last updated on 07/Mar/24
$${Do}\:{you}\:{mean}\:\:\left(\mathrm{A}\cap\mathrm{C}\right)\:\cup\:\left(\mathrm{A}\cap\mathrm{B}\right)\:\cup\:\left(\mathrm{B}\cap\mathrm{C}\right)\:\:? \\ $$
Commented by cortano12 last updated on 07/Mar/24
$$\mathrm{no} \\ $$
Commented by A5T last updated on 07/Mar/24
$$“+''\:{generally}\:{is}\:{used}\:{to}\:{sum}\:{cardinality}\:{of}\:{sets}. \\ $$$${Are}\:{you}\:{referring}\:{to}\:{that}? \\ $$
Commented by cortano12 last updated on 07/Mar/24
$$\mathrm{A}+\mathrm{B}\:=\:\left(\mathrm{A}\cup\mathrm{B}\right)−\left(\mathrm{A}\cap\mathrm{B}\right)\: \\ $$$$\:\mathrm{or}\:\mathrm{A}+\mathrm{B}\:=\:\left(\mathrm{A}−\mathrm{B}\right)\:\cup\:\left(\mathrm{B}−\mathrm{A}\right)\: \\ $$
Commented by A5T last updated on 07/Mar/24
$${Then}\:{D}=\left({A}\cap{C}\right)+\left({A}\cap{B}\right)=\left\{{a},{c}\right\} \\ $$$${D}\:+\:\left({B}\cap{C}\right)=\left({A}\cap{C}\right)+\left({A}\cap{B}\right)+\left({B}\cap{C}\right)=\left\{{a},{b},{c},{d}\right\} \\ $$