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Question-53043




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

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