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Question Number 114233 by Aina Samuel Temidayo last updated on 18/Sep/20

$$\mathrm{If}\mathrm{two}\mathrm{sets}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{are}\mathrm{having}99$$$$\mathrm{elements}\mathrm{in}\mathrm{common},\mathrm{then}\mathrm{the}$$$$\mathrm{number}\mathrm{of}\mathrm{elements}\mathrm{common}\mathrm{to}\mathrm{each}$$$$\mathrm{of}\mathrm{the}\mathrm{sets}\mathrm{A}\times \mathrm{B}\mathrm{and}\mathrm{B}\times \mathrm{A}\mathrm{is}$$

Answered by 1549442205PVT last updated on 20/Sep/20

$$\mathrm{A}=\{{\mathrm{x}}_1,{\mathrm{x}}_2,...,{\mathrm{x}}_{\mathrm{m}},{\mathrm{c}}_1,{\mathrm{c}}_2,...,{\mathrm{c}}_{99}\}$$$$\mathrm{B}=\{{\mathrm{y}}_1,{\mathrm{y}}_2,...,{\mathrm{y}}_{\mathrm{n}},{\mathrm{c}}_1,{\mathrm{c}}_2,...,{\mathrm{c}}_{99}\}$$$$\Rightarrow \mathrm{A}\times \mathrm{B}\mathrm{and}\mathrm{B}\times \mathrm{A}{\mathrm{haveF}}_{99}^2={99}^2$$$$=9801\mathrm{common}\mathrm{elements}\mathrm{because}$$$$\mathrm{the}\mathrm{common}\mathrm{elements}\mathrm{are}$$$$\left\{({\mathrm{c}}_{\mathrm{i}},{\mathrm{c}}_{\mathrm{j}})(1⩽\mathrm{i},\mathrm{j}⩽99)\right\}$$

Commented by 1549442205PVT last updated on 18/Sep/20

$$\mathrm{Thank}\mathrm{you}.\mathrm{I}\mathrm{corrected}.\mathrm{You}\mathrm{just}\mathrm{need}$$$$\mathrm{instead}\mathrm{of}99\mathrm{by}4\mathrm{and}\mathrm{check}$$

Commented by Aina Samuel Temidayo last updated on 18/Sep/20

$$\mathrm{Not}\mathrm{correct}.$$

Commented by 1549442205PVT last updated on 20/Sep/20

$$\mathrm{It}\mathrm{is}{\mathrm{F}}_{99}^2.\mathrm{Note}{\mathrm{F}}_{\mathrm{n}}^{\mathrm{k}}={\mathrm{n}}^{\mathrm{k}}.\mathrm{Note}\mathrm{that}$$$$\mathrm{Given}\mathrm{two}\mathrm{sets}\mathrm{A}={\left\{{\mathrm{x}}_{\mathrm{i}}\right\}}_{\mathrm{i}=1}^{\mathrm{m}},\mathrm{B}={\left\{{\mathrm{y}}_{\mathrm{j}}\right\}}_{\mathrm{j}=1}^{\mathrm{n}}$$$$\mathrm{Then}\mathrm{A}\times \mathrm{B}=\{({\mathrm{x}}_{\mathrm{i}},{\mathrm{y}}_{\mathrm{j}})\mid 1⩽\mathrm{i}⩽\mathrm{m},1⩽\mathrm{j}⩽\mathrm{n}\}$$$$\mathrm{B}\times \mathrm{A}=\{({\mathrm{y}}_{\mathrm{i}},{\mathrm{x}}_{\mathrm{j}})\mid 1⩽\mathrm{i}⩽\mathrm{n},1⩽\mathrm{j}⩽\mathrm{m}\}$$

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