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Question Number 227151 by efronzo1 last updated on 03/Jan/26
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Answered by mr W last updated on 04/Jan/26
$${she}\:{can}\:{choose}\:{numbers}\:{from}\:\mathrm{1}\:{to}\:\mathrm{17}. \\ $$$${these}\:{numbers}\:{can}\:{be}\:{arranged}\:{in} \\ $$$${following}\:\mathrm{8}\:{groups}: \\ $$$$\mathrm{1},\:\mathrm{17} \\ $$$$\mathrm{2},\:\mathrm{16} \\ $$$$\mathrm{3},\:\mathrm{15} \\ $$$$\mathrm{4},\:\mathrm{14} \\ $$$$\mathrm{5},\:\mathrm{13} \\ $$$$\mathrm{6},\:\mathrm{12} \\ $$$$\mathrm{7},\:\mathrm{11} \\ $$$$\mathrm{8},\:\mathrm{10} \\ $$$${the}\:{number}\:\mathrm{9}\:{remains}. \\ $$$${since}\:{the}\:{sum}\:{of}\:{the}\:{numbers}\:{in} \\ $$$${a}\:{group}\:{is}\:\mathrm{18},\:{she}\:{can}\:{only}\:{choose} \\ $$$${at}\:{most}\:{one}\:{number}\:{from}\:{each} \\ $$$${group}.\:{in}\:{this}\:{way}\:{she}\:{can}\:{choose} \\ $$$${at}\:{most}\:\mathrm{8}\:{numbers}.\:{the}\:\mathrm{9}^{{th}} \:{number} \\ $$$${must}\:{be}\:{the}\:{remaining}\:{number}\:\mathrm{9}. \\ $$$${that}\:{means}\:{the}\:{number}\:\mathrm{9}\:{must}\:{be} \\ $$$${always}\:{included}.\underbrace{ } \\ $$