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-

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

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

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

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

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}

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

Facebook Tweet Pin

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-

Leave a Reply Cancel reply