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Q.If A×B={(a,1),(a,5),(1,2),(b,2),(b,5),(b,1)}A \times B = \{(a,1), (a,5), (1,2), (b,2), (b,5), (b,1)\} find B×AB \times A.

Manipur CohsemCouncil of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2021Subjective· 1mImportance★★★★★
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Reading the printed set as A×B={(a,1),(a,2),(a,5),(b,1),(b,2),(b,5)}A\times B=\{(a,1),(a,2),(a,5),(b,1),(b,2),(b,5)\} (correcting a scan misprint of "(a,2)" as "(1,2)"), A={a,b}A=\{a,b\} and B={1,2,5}B=\{1,2,5\}, giving B×A={(1,a),(1,b),(2,a),(2,b),(5,a),(5,b)}B\times A=\{(1,a),(1,b),(2,a),(2,b),(5,a),(5,b)\}.

Noting a likely scan issue: the printed set {(a,1),(a,5),(1,2),(b,2),(b,5),(b,1)}\{(a,1),(a,5),(1,2),(b,2),(b,5),(b,1)\} has one pair, (1,2)(1,2), whose first coordinate breaks the pattern — in a genuine Cartesian product A×BA\times B every first coordinate must come from the same set AA, and here "bb" already appears paired with all three of 1,2,51,2,5 while "aa" appears with only 1,51,5 (missing 22). This is consistent with "(1,2)" being a scanning misread of "(a,2)" (the two characters look similar in some fonts). With that one correction the six pairs read as a complete, consistent product:

A×B={(a,1),(a,2),(a,5),(b,1),(b,2),(b,5)}.A\times B=\{(a,1),(a,2),(a,5),(b,1),(b,2),(b,5)\}.

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