Question and Answers Forum

All Questions      Topic List

Set Theory Questions

Previous in All Question      Next in All Question      

Previous in Set Theory      Next in Set Theory      

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

If two sets A and B are having 99 elements in common, then the number of elements common to each of the sets A×B and B×A is

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

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

A={x_1 ,x_2 ,...,x_m ,c_1 ,c_2 ,...,c_(99) } B={y_1 ,y_2 ,...,y_n ,c_1 ,c_2 ,...,c_(99) } ⇒A×B and B×A haveF_(99) ^2 =99^2 =9801common elements because the common elements are {(c_i ,c_j )(1≤i,j≤99)}

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

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

Thank you .I corrected.You just need instead of 99 by 4 and check

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

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

Not correct.

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

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

It is F_(99) ^2 .Note F_n ^k =n^k .Note that Given two sets A={x_i }_(i=1) ^m ,B={y_j }_(j=1) ^n Then A×B={(x_i ,y_j )∣1≤i≤m,1≤j≤n} B×A={(y_i ,x_j )∣1≤i≤n,1≤j≤m}

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com