Question Number 53043 by Tinkutara last updated on 16/Jan/19
Commented by tanmay.chaudhury50@gmail.com last updated on 17/Jan/19
$${excellent}\:_{} {question}… \\ $$
Answered by mr W last updated on 17/Jan/19
$${let}'{s}\:{say}\:{the}\:{two}\:{numbers}\:{are}\:{X}\:{and}\:{Y} \\ $$$${with}\:{X}<{Y} \\ $$$${for}\:{each}\:{X}\:{in}\:\mathrm{1}−\mathrm{90}\:{there}\:{are}\:\mathrm{10}\:{Ys} \\ $$$$\Rightarrow{totally}\:\mathrm{90}×\mathrm{10}=\mathrm{900} \\ $$$${for}\:{X}=\mathrm{91}:\:\mathrm{9}\:{Ys} \\ $$$${for}\:{X}=\mathrm{92}:\:\mathrm{8}\:{Ys} \\ $$$${for}\:{X}=\mathrm{93}:\:\mathrm{7}\:{Ys} \\ $$$$… \\ $$$${for}\:{X}=\mathrm{99}:\:\mathrm{1}\:{Y} \\ $$$$\Rightarrow\mathrm{9}+\mathrm{8}+…+\mathrm{1}=\mathrm{45} \\ $$$$ \\ $$$$\mathrm{900}+\mathrm{45}=\mathrm{945} \\ $$$${i}.{e}.\:{there}\:{are}\:\mathrm{945}\:{ways}. \\ $$$$ \\ $$$$\left({a}\right)\left({b}\right)\left({c}\right)\:{are}\:{all}\:{correct}. \\ $$
Commented by Tinkutara last updated on 18/Jan/19
Thank you Sir!
Answered by tanmay.chaudhury50@gmail.com last updated on 18/Jan/19
$${difference}=\mathrm{1} \\ $$$$\left(\mathrm{1},\mathrm{2}\right) \\ $$$$\left(\mathrm{2},\mathrm{3}\right)\:\:\:\:\:\:\mathrm{99}{c}_{\mathrm{1}} \\ $$$$… \\ $$$$…\left(\mathrm{99},\mathrm{100}\right) \\ $$$${dif}=\mathrm{2} \\ $$$$\left(\mathrm{1},\mathrm{3}\right) \\ $$$$\mathrm{2},\mathrm{4} \\ $$$$\mathrm{3},\mathrm{5} \\ $$$$…\:\:\:\:\:\:\:\:\mathrm{98}{c}_{\mathrm{1}} \\ $$$$… \\ $$$$\mathrm{98},\mathrm{100} \\ $$$${dif}=\mathrm{3} \\ $$$$\left(\mathrm{1},\mathrm{4}\right) \\ $$$$\left(\mathrm{2},\mathrm{5}\right) \\ $$$$\left(\mathrm{3},\mathrm{6}\right) \\ $$$$…\:\:\:\:\:\:\mathrm{97}{c}_{\mathrm{1}} \\ $$$$… \\ $$$$\left(\mathrm{97},\mathrm{100}\right) \\ $$$$ \\ $$$${so}\:{answer}\:{is} \\ $$$$\mathrm{99}{c}_{\mathrm{1}} +\mathrm{98}{c}_{\mathrm{1}} +\mathrm{97}{c}_{\mathrm{1}} +\mathrm{96}{c}_{\mathrm{1}} +\mathrm{95}{c}_{\mathrm{1}} +\mathrm{94}{c}_{\mathrm{1}} +\mathrm{93}{c}_{\mathrm{1}} +\mathrm{92}{c}_{\mathrm{1}} \\ $$$$+\mathrm{91}{c}_{\mathrm{1}} +\mathrm{90}{c}_{\mathrm{1}} \\ $$$$=\mathrm{99}+\mathrm{98}+\mathrm{97}+…+\mathrm{90} \\ $$$$=\frac{\mathrm{10}}{\mathrm{2}}\left[\mathrm{2}×\mathrm{99}+\left(\mathrm{10}−\mathrm{1}\right)×−\mathrm{1}\right] \\ $$$$=\mathrm{5}\left(\mathrm{198}−\mathrm{9}\right)=\mathrm{5}×\mathrm{189}=\mathrm{945} \\ $$$$ \\ $$$$ \\ $$
Commented by Tinkutara last updated on 18/Jan/19
Thank you Sir!