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Question Number 180335 by SLVR last updated on 10/Nov/22
The integers between 1 to 10^4   contain exactly one 8  and one 9  is? I got 1160 but one  is arguing  1154only..kindly help me out
$${The}\:{integers}\:{between}\:\mathrm{1}\:{to}\:\mathrm{10}^{\mathrm{4}} \\ $$$${contain}\:{exactly}\:{one}\:\mathrm{8}\:\:{and}\:{one}\:\mathrm{9} \\ $$$${is}?\:{I}\:{got}\:\mathrm{1160}\:{but}\:{one}\:\:{is}\:{arguing} \\ $$$$\mathrm{1154}{only}..{kindly}\:{help}\:{me}\:{out} \\ $$
Commented by mr W last updated on 10/Nov/22
i think both are wrong, or i didn′t  understand the question correctly.
$${i}\:{think}\:{both}\:{are}\:{wrong},\:{or}\:{i}\:{didn}'{t} \\ $$$${understand}\:{the}\:{question}\:{correctly}. \\ $$
Answered by mr W last updated on 10/Nov/22
1) 2 digit numbers:  89 and 98 ⇒2 numbers    2) 3 digit numbers:  Y89, 8X9, 89X   ⇒ 2×(7+8+8)=46 numbers  X=0,1,2,...,7   Y=1,2,...,7    3) 4 digit numbers:  YX89 ⇒2×7×8=112  Y8X9 ⇒2×7×8  Y89X ⇒2×7×8  8XX9 ⇒2×8×8=128  8X9X ⇒2×8×8  89XX ⇒2×8×8  ⇒3×112+3×128=720    totally: 2+46+720=768
$$\left.\mathrm{1}\right)\:\mathrm{2}\:{digit}\:{numbers}: \\ $$$$\mathrm{89}\:{and}\:\mathrm{98}\:\Rightarrow\mathrm{2}\:{numbers} \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:\mathrm{3}\:{digit}\:{numbers}: \\ $$$${Y}\mathrm{89},\:\mathrm{8}{X}\mathrm{9},\:\mathrm{89}{X}\: \\ $$$$\Rightarrow\:\mathrm{2}×\left(\mathrm{7}+\mathrm{8}+\mathrm{8}\right)=\mathrm{46}\:{numbers} \\ $$$${X}=\mathrm{0},\mathrm{1},\mathrm{2},…,\mathrm{7}\: \\ $$$${Y}=\mathrm{1},\mathrm{2},…,\mathrm{7} \\ $$$$ \\ $$$$\left.\mathrm{3}\right)\:\mathrm{4}\:{digit}\:{numbers}: \\ $$$${YX}\mathrm{89}\:\Rightarrow\mathrm{2}×\mathrm{7}×\mathrm{8}=\mathrm{112} \\ $$$${Y}\mathrm{8}{X}\mathrm{9}\:\Rightarrow\mathrm{2}×\mathrm{7}×\mathrm{8} \\ $$$${Y}\mathrm{89}{X}\:\Rightarrow\mathrm{2}×\mathrm{7}×\mathrm{8} \\ $$$$\mathrm{8}{XX}\mathrm{9}\:\Rightarrow\mathrm{2}×\mathrm{8}×\mathrm{8}=\mathrm{128} \\ $$$$\mathrm{8}{X}\mathrm{9}{X}\:\Rightarrow\mathrm{2}×\mathrm{8}×\mathrm{8} \\ $$$$\mathrm{89}{XX}\:\Rightarrow\mathrm{2}×\mathrm{8}×\mathrm{8} \\ $$$$\Rightarrow\mathrm{3}×\mathrm{112}+\mathrm{3}×\mathrm{128}=\mathrm{720} \\ $$$$ \\ $$$${totally}:\:\mathrm{2}+\mathrm{46}+\mathrm{720}=\mathrm{768} \\ $$
Commented by SLVR last updated on 11/Nov/22
Thanks...Prof.W..really i counted  repeatedly..i am wrong sir...
$${Thanks}…{Prof}.{W}..{really}\:{i}\:{counted} \\ $$$${repeatedly}..{i}\:{am}\:{wrong}\:{sir}… \\ $$
Answered by Acem last updated on 10/Nov/22
You have three general cases   2 dig         3 dig       4 dig    89              x89          xx89   with 2! permutation^(8,9)      x89       3! = 3 ×2^(8,9)    xx89    ((4!)/(2!))= 12 = 6 ×2^(8,9)      well let′s note cases  • x′89 , 8x9 , 89x ∣_( x: 0→ 7) ^(x′: 1→ 7)      • x′x89, x′8x9 , x′89x ∣_(x′: 1→7) ^(x: 0→7)       8xx9, 8x9x , 89xx^(x: 0→7)      Num_(Numb.) = 2 [1+(7+2×8)+3(7×8+8×8)]                         = 2[1+23+24(15)]= 768 numbers
$${You}\:{have}\:{three}\:{general}\:{cases} \\ $$$$\:\mathrm{2}\:{dig}\:\:\:\:\:\:\:\:\:\mathrm{3}\:{dig}\:\:\:\:\:\:\:\mathrm{4}\:{dig} \\ $$$$\:\:\mathrm{89}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\mathrm{89}\:\:\:\:\:\:\:\:\:\:{xx}\mathrm{89}\:\:\:{with}\:\mathrm{2}!\:{permutation}^{\mathrm{8},\mathrm{9}} \\ $$$$ \\ $$$$\:{x}\mathrm{89}\:\:\:\:\:\:\:\mathrm{3}!\:=\:\mathrm{3}\:×\mathrm{2}^{\mathrm{8},\mathrm{9}} \\ $$$$\:{xx}\mathrm{89}\:\:\:\:\frac{\mathrm{4}!}{\mathrm{2}!}=\:\mathrm{12}\:=\:\mathrm{6}\:×\mathrm{2}^{\mathrm{8},\mathrm{9}} \:\:\: \\ $$$${well}\:{let}'{s}\:{note}\:{cases} \\ $$$$\bullet\:{x}'\mathrm{89}\:,\:\mathrm{8}{x}\mathrm{9}\:,\:\mathrm{89}{x}\:\mid_{\:{x}:\:\mathrm{0}\rightarrow\:\mathrm{7}} ^{{x}':\:\mathrm{1}\rightarrow\:\mathrm{7}} \: \\ $$$$ \\ $$$$\bullet\:{x}'{x}\mathrm{89},\:{x}'\mathrm{8}{x}\mathrm{9}\:,\:{x}'\mathrm{89}{x}\:\mid_{{x}':\:\mathrm{1}\rightarrow\mathrm{7}} ^{{x}:\:\mathrm{0}\rightarrow\mathrm{7}} \\ $$$$\:\:\:\:\mathrm{8}{xx}\mathrm{9},\:\mathrm{8}{x}\mathrm{9}{x}\:,\:\mathrm{89}{xx}\:^{{x}:\:\mathrm{0}\rightarrow\mathrm{7}} \\ $$$$ \\ $$$$\:{Num}_{{Numb}.} =\:\mathrm{2}\:\left[\mathrm{1}+\left(\mathrm{7}+\mathrm{2}×\mathrm{8}\right)+\mathrm{3}\left(\mathrm{7}×\mathrm{8}+\mathrm{8}×\mathrm{8}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\left[\mathrm{1}+\mathrm{23}+\mathrm{24}\left(\mathrm{15}\right)\right]=\:\mathrm{768}\:{numbers} \\ $$$$ \\ $$

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