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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=1 (1,2) (2,3) 99c_1 ... ...(99,100) dif=2 (1,3) 2,4 3,5 ... 98c_1 ... 98,100 dif=3 (1,4) (2,5) (3,6) ... 97c_1 ... (97,100) so answer is 99c_1 +98c_1 +97c_1 +96c_1 +95c_1 +94c_1 +93c_1 +92c_1 +91c_1 +90c_1 =99+98+97+...+90 =((10)/2)[2×99+(10−1)×−1] =5(198−9)=5×189=945](Q53117.png)
$${difference}=\mathrm{1} \\ $$$$\left(\mathrm{1},\mathrm{2}\right) \\ $$$$\left(\mathrm{2},\mathrm{3}\right)\:\:\:\:\:\:\mathrm{99}{c}_{\mathrm{1}} \\ $$$$... \\ $$$$...\left(\mathrm{99},\mathrm{100}\right) \\ $$$${dif}=\mathrm{2} \\ $$$$\left(\mathrm{1},\mathrm{3}\right) \\ $$$$\mathrm{2},\mathrm{4} \\ $$$$\mathrm{3},\mathrm{5} \\ $$$$...\:\:\:\:\:\:\:\:\mathrm{98}{c}_{\mathrm{1}} \\ $$$$... \\ $$$$\mathrm{98},\mathrm{100} \\ $$$${dif}=\mathrm{3} \\ $$$$\left(\mathrm{1},\mathrm{4}\right) \\ $$$$\left(\mathrm{2},\mathrm{5}\right) \\ $$$$\left(\mathrm{3},\mathrm{6}\right) \\ $$$$...\:\:\:\:\:\:\mathrm{97}{c}_{\mathrm{1}} \\ $$$$... \\ $$$$\left(\mathrm{97},\mathrm{100}\right) \\ $$$$ \\ $$$${so}\:{answer}\:{is} \\ $$$$\mathrm{99}{c}_{\mathrm{1}} +\mathrm{98}{c}_{\mathrm{1}} +\mathrm{97}{c}_{\mathrm{1}} +\mathrm{96}{c}_{\mathrm{1}} +\mathrm{95}{c}_{\mathrm{1}} +\mathrm{94}{c}_{\mathrm{1}} +\mathrm{93}{c}_{\mathrm{1}} +\mathrm{92}{c}_{\mathrm{1}} \\ $$$$+\mathrm{91}{c}_{\mathrm{1}} +\mathrm{90}{c}_{\mathrm{1}} \\ $$$$=\mathrm{99}+\mathrm{98}+\mathrm{97}+...+\mathrm{90} \\ $$$$=\frac{\mathrm{10}}{\mathrm{2}}\left[\mathrm{2}×\mathrm{99}+\left(\mathrm{10}−\mathrm{1}\right)×−\mathrm{1}\right] \\ $$$$=\mathrm{5}\left(\mathrm{198}−\mathrm{9}\right)=\mathrm{5}×\mathrm{189}=\mathrm{945} \\ $$$$ \\ $$$$ \\ $$
Commented by Tinkutara last updated on 18/Jan/19
Thank you Sir!