Question-214536

Question Number 214536 by ajfour last updated on 11/Dec/24

Commented by ajfour last updated on 11/Dec/24

Out of questions labelled by just permutations of digits of the number 123456 what rank does this Q hold? In increasing order, that is.

$${Out}\:{of}\:{questions}\:{labelled}\:{by}\:{just} \\ $$$${permutations}\:{of}\:{digits}\:{of}\:{the}\:{number} \\ $$$$\mathrm{123456}\:\:\:{what}\:{rank}\:{does}\:{this}\:{Q}\:{hold}? \\ $$$${In}\:{increasing}\:{order},\:{that}\:{is}. \\ $$

Commented by A5T last updated on 11/Dec/24

x_1 x_2 x_3 x_4 x_5 x_6 ^(______________) when x_1 =1, there are 5!=120 permutations of (x_2 ,x_3 ,x_4 ,x_5 ,x_6 ) when x_1 =2 and x_2 =1 for x_3 =3; there are 3!=6 for (4,5,6) ⇒214356 − (120+6+1)th ⇒214365 − 128th ⇒214536 − 129th

$$\overset{\_\_\_\_\_\_\_\_\_\_\_\_\_\_} {{x}_{\mathrm{1}} {x}_{\mathrm{2}} {x}_{\mathrm{3}} {x}_{\mathrm{4}} {x}_{\mathrm{5}} {x}_{\mathrm{6}} }\: \\ $$$${when}\:{x}_{\mathrm{1}} =\mathrm{1},\:{there}\:{are}\:\mathrm{5}!=\mathrm{120}\:{permutations}\:{of}\: \\ $$$$\left({x}_{\mathrm{2}} ,{x}_{\mathrm{3}} ,{x}_{\mathrm{4}} ,{x}_{\mathrm{5}} ,{x}_{\mathrm{6}} \right) \\ $$$${when}\:{x}_{\mathrm{1}} =\mathrm{2}\:{and}\:{x}_{\mathrm{2}} =\mathrm{1} \\ $$$${for}\:{x}_{\mathrm{3}} =\mathrm{3};\:{there}\:{are}\:\mathrm{3}!=\mathrm{6}\:{for}\:\left(\mathrm{4},\mathrm{5},\mathrm{6}\right) \\ $$$$\Rightarrow\mathrm{214356}\:−\:\left(\mathrm{120}+\mathrm{6}+\mathrm{1}\right){th} \\ $$$$\Rightarrow\mathrm{214365}\:−\:\mathrm{128}{th} \\ $$$$\Rightarrow\mathrm{214536}\:−\:\mathrm{129}{th} \\ $$

Answered by A5T last updated on 11/Dec/24

Commented by A5T last updated on 11/Dec/24

(BD−1)^2 =1×3⇒BD=1+(√3) AD=AE=1⇒AB=1+1+(√3)=2+(√3) (EC−r)^2 =r(r+2)⇒EC=r+(√(r(r+2))) ⇒AC=1+r+(√(r(r+2))) ((AB×AC)/2)=((1(AB+AC+(√(AB^2 +AC^2 ))))/2) ⇒(2+(√3))(1+r+(√(r(r+2)))) =2+(√3)+1+r+(√(r(r+2)))+(√(7+4(√3)+1+r^2 +r(r+2)+2r+2(√(r(r+2)))+2r(√(r(r+2))))) ⇒r=((9+5(√3))/(24))

$$\left({BD}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1}×\mathrm{3}\Rightarrow{BD}=\mathrm{1}+\sqrt{\mathrm{3}} \\ $$$${AD}={AE}=\mathrm{1}\Rightarrow{AB}=\mathrm{1}+\mathrm{1}+\sqrt{\mathrm{3}}=\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$$\left({EC}−{r}\right)^{\mathrm{2}} ={r}\left({r}+\mathrm{2}\right)\Rightarrow{EC}={r}+\sqrt{{r}\left({r}+\mathrm{2}\right)} \\ $$$$\Rightarrow{AC}=\mathrm{1}+{r}+\sqrt{{r}\left({r}+\mathrm{2}\right)} \\ $$$$\frac{{AB}×{AC}}{\mathrm{2}}=\frac{\mathrm{1}\left({AB}+{AC}+\sqrt{{AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} }\right)}{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\left(\mathrm{1}+{r}+\sqrt{{r}\left({r}+\mathrm{2}\right)}\right) \\ $$$$=\mathrm{2}+\sqrt{\mathrm{3}}+\mathrm{1}+{r}+\sqrt{{r}\left({r}+\mathrm{2}\right)}+\sqrt{\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{1}+{r}^{\mathrm{2}} +{r}\left({r}+\mathrm{2}\right)+\mathrm{2}{r}+\mathrm{2}\sqrt{{r}\left({r}+\mathrm{2}\right)}+\mathrm{2}{r}\sqrt{{r}\left({r}+\mathrm{2}\right)}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{9}+\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{24}} \\ $$

Commented by ajfour last updated on 11/Dec/24

$${Awesome}! \\ $$

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