Question Number 242 by 123456 last updated on 25/Jan/15
$$\mathrm{give}\:\mathrm{the}\:\mathrm{sets} \\ $$$$\mathrm{A}=\left\{{x}\in\mathbb{N}:\mathrm{0}\leqslant{x}\leqslant\mathrm{9}\right\} \\ $$$$\mathrm{B}=\left\{\mathrm{0}\right\} \\ $$$$\mathrm{C}=\left\{\mathrm{1}\right\} \\ $$$$\mathrm{D}=\left\{\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{7}\right\} \\ $$$$\mathrm{E}=\left\{\mathrm{4},\mathrm{6},\mathrm{8},\mathrm{9}\right\} \\ $$$$\mathrm{compute} \\ $$$$\mid\mathrm{A}\backslash\mathrm{B}\mid+\mid\mathrm{A}\backslash\mathrm{C}\mid+\mid\mathrm{A}\backslash\mathrm{D}\mid+\mid\mathrm{A}\backslash\mathrm{E}\mid \\ $$$$\mathrm{where} \\ $$$$\mid\mathrm{X}\mid=\mathrm{number}\:\mathrm{of}\:\mathrm{elements}\:\mathrm{of}\:\mathrm{X} \\ $$$$\mathrm{X}\backslash\mathrm{Y}=\left\{{x}:{x}\in\mathrm{X}\:\boldsymbol{\mathrm{e}}\:{x}\notin\mathrm{Y}\right\} \\ $$$$\mathrm{X}−\mathrm{Y}=\left\{{x}:{x}\in\mathrm{X}\:\boldsymbol{\mathrm{e}}\:{x}\notin\mathrm{Y}\right\} \\ $$
Answered by prakash jain last updated on 17/Dec/14
$$\mid\mathrm{A}\backslash\mathrm{B}\mid=\mathrm{9} \\ $$$$\mid\mathrm{A}\backslash\mathrm{C}\mid=\mathrm{9} \\ $$$$\mid\mathrm{A}\backslash\mathrm{D}\mid=\mathrm{6} \\ $$$$\mid\mathrm{A}\backslash\mathrm{E}\mid=\mathrm{6} \\ $$$$\mid\mathrm{A}\backslash\mathrm{B}\mid+\mid\mathrm{A}\backslash\mathrm{C}\mid+\mid\mathrm{A}\backslash\mathrm{D}\mid+\mid\mathrm{A}\backslash\mathrm{E}\mid=\mathrm{30} \\ $$