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Question Number 67281 by Tony Lin last updated on 25/Aug/19

Commented by mr W last updated on 25/Aug/19

I get 3C_2 ^5 +3C_2 ^5 +12=72

$${I}\:{get}\:\mathrm{3}{C}_{\mathrm{2}} ^{\mathrm{5}} +\mathrm{3}{C}_{\mathrm{2}} ^{\mathrm{5}} +\mathrm{12}=\mathrm{72} \\ $$

Commented by Tony Lin last updated on 25/Aug/19

Repost Question No.66681

$${Repost}\:{Question}\:{No}.\mathrm{66681} \\ $$

Commented by Tony Lin last updated on 25/Aug/19

C_2 ^3 C_2 ^5 ×3−C_2 ^3 C_2 ^3 ×2=72

$${C}_{\mathrm{2}} ^{\mathrm{3}} {C}_{\mathrm{2}} ^{\mathrm{5}} ×\mathrm{3}−{C}_{\mathrm{2}} ^{\mathrm{3}} {C}_{\mathrm{2}} ^{\mathrm{3}} ×\mathrm{2}=\mathrm{72} \\ $$

Commented by Tony Lin last updated on 25/Aug/19

Commented by Tony Lin last updated on 25/Aug/19

orange part(2×4 inside) ⇒choose two from three vertical lines ⇒choose two from five horizontal lines ⇒C_2 ^3 C_2 ^5 =30 blue part(2×4 outside) ⇒the same as the orange part ⇒C_2 ^3 C_2 ^5 =30 green part(2×4 middle) ⇒only in this part the rectangles outside combine with the rectangles inside ⇒the same as the orange part ⇒C_2 ^3 C_2 ^5 =30 pink part(2×2 inside) ⇒we have already counted this part in the orange part but we count this part again in the green part ⇒choose two from three vertical lines ⇒choose two from three horizontal lines ⇒C_2 ^3 C_2 ^3 =9 yellow part(2×2 outside) ⇒we have already counted this part in the blue part but we count this part again in the green part ⇒the same as the pink part ⇒C_2 ^3 C_2 ^3 =9 ∴there are (C_2 ^5 C_2 ^3 ×3−C_2 ^3 C_2 ^3 ×2=72) rectangles

$${orange}\:{part}\left(\mathrm{2}×\mathrm{4}\:{inside}\right) \\ $$$$\Rightarrow{choose}\:{two}\:{from}\:{three}\:{vertical}\:{lines} \\ $$$$\Rightarrow{choose}\:{two}\:{from}\:{five}\:{horizontal}\:{lines} \\ $$$$\Rightarrow{C}_{\mathrm{2}} ^{\mathrm{3}} {C}_{\mathrm{2}} ^{\mathrm{5}} =\mathrm{30} \\ $$$${blue}\:{part}\left(\mathrm{2}×\mathrm{4}\:{outside}\right) \\ $$$$\Rightarrow{the}\:{same}\:{as}\:{the}\:{orange}\:{part} \\ $$$$\Rightarrow{C}_{\mathrm{2}} ^{\mathrm{3}} {C}_{\mathrm{2}} ^{\mathrm{5}} =\mathrm{30} \\ $$$${green}\:{part}\left(\mathrm{2}×\mathrm{4}\:{middle}\right) \\ $$$$\Rightarrow{only}\:{in}\:{this}\:{part}\:{the}\:{rectangles}\:{outside} \\ $$$${combine}\:{with}\:\:{the}\:{rectangles}\:{inside}\: \\ $$$$\Rightarrow{the}\:{same}\:{as}\:{the}\:{orange}\:{part} \\ $$$$\Rightarrow{C}_{\mathrm{2}} ^{\mathrm{3}} {C}_{\mathrm{2}} ^{\mathrm{5}} =\mathrm{30} \\ $$$${pink}\:{part}\left(\mathrm{2}×\mathrm{2}\:{inside}\right) \\ $$$$\Rightarrow{we}\:{have}\:{already}\:{counted}\:{this}\:{part} \\ $$$${in}\:{the}\:{orange}\:{part}\:{but}\:{we}\:{count}\:{this}\: \\ $$$${part}\:{again}\:{in}\:{the}\:{green}\:{part} \\ $$$$\Rightarrow{choose}\:{two}\:{from}\:{three}\:{vertical}\:{lines} \\ $$$$\Rightarrow{choose}\:{two}\:{from}\:{three}\:{horizontal}\:{lines} \\ $$$$\Rightarrow{C}_{\mathrm{2}} ^{\mathrm{3}} {C}_{\mathrm{2}} ^{\mathrm{3}} =\mathrm{9} \\ $$$${yellow}\:{part}\left(\mathrm{2}×\mathrm{2}\:{outside}\right) \\ $$$$\Rightarrow{we}\:{have}\:{already}\:{counted}\:{this}\:{part} \\ $$$${in}\:{the}\:{blue}\:{part}\:{but}\:{we}\:{count}\:{this} \\ $$$${part}\:{again}\:{in}\:{the}\:{green}\:{part} \\ $$$$\Rightarrow{the}\:{same}\:{as}\:{the}\:{pink}\:{part} \\ $$$$\Rightarrow{C}_{\mathrm{2}} ^{\mathrm{3}} {C}_{\mathrm{2}} ^{\mathrm{3}} =\mathrm{9} \\ $$$$\therefore{there}\:{are}\:\left({C}_{\mathrm{2}} ^{\mathrm{5}} {C}_{\mathrm{2}} ^{\mathrm{3}} ×\mathrm{3}−{C}_{\mathrm{2}} ^{\mathrm{3}} {C}_{\mathrm{2}} ^{\mathrm{3}} ×\mathrm{2}=\mathrm{72}\right)\:{rectangles} \\ $$

Commented by Rasheed.Sindhi last updated on 25/Aug/19

Sir Tony Lin, I don′t see any error in your approach.However I want to see where′s deffect in my approach! Pl cooperate with me.

$$\mathrm{Sir}\:\mathrm{Tony}\:\mathrm{Lin},\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{see}\:\mathrm{any}\:\mathrm{error}\:\mathrm{in} \\ $$$$\mathrm{your}\:\mathrm{approach}.\mathrm{However}\:\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{see} \\ $$$$\mathrm{where}'\mathrm{s}\:\mathrm{deffect}\:\mathrm{in}\:\mathrm{my}\:\mathrm{approach}!\:\mathrm{Pl} \\ $$$$\mathrm{cooperate}\:\mathrm{with}\:\mathrm{me}. \\ $$

Answered by Rasheed.Sindhi last updated on 25/Aug/19

Commented by Rasheed.Sindhi last updated on 25/Aug/19

^• In order to prevent missing/repeating in counting,we′ll count rectangles with their left-upper corners.For example all the rectangles whose left-upper corner is A will be counted together and recorded with A.In this way we will count rectangles with each and every vertex. ^• A: Edges from A to the left: AB,AC (2 edges) Edges from A to the down: AG,AL,AQ,AA_1 (4 edges) ∴Number of rectangles with left- upper corner A=2×4=8 (Recall that every upper edge makes a rectangle with every left edge.) .......

$$\:^{\bullet} {In}\:{order}\:{to}\:{prevent}\:{missing}/{repeating} \\ $$$${in}\:{counting},{we}'{ll}\:{count}\:{rectangles}\:{with} \\ $$$${their}\:\boldsymbol{{left}}-\boldsymbol{{upper}}\:\boldsymbol{{corners}}.{For}\:{example} \\ $$$${all}\:{the}\:{rectangles}\:{whose}\:{left}-{upper} \\ $$$${corner}\:{is}\:\mathrm{A}\:{will}\:{be}\:{counted}\:{together} \\ $$$${and}\:{recorded}\:{with}\:\mathrm{A}.{In}\:{this}\:{way}\:{we} \\ $$$${will}\:{count}\:{rectangles}\:{with}\:{each}\:{and} \\ $$$${every}\:{vertex}. \\ $$$$\:^{\bullet} \mathrm{A}:\:\mathcal{E}{dges}\:{from}\:\mathrm{A}\:{to}\:{the}\:{left}: \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{AB},\mathrm{AC}\:\left(\mathrm{2}\:{edges}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\mathcal{E}{dges}\:{from}\:\mathrm{A}\:{to}\:{the}\:{down}: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{AG},\mathrm{AL},\mathrm{AQ},\mathrm{AA}_{\mathrm{1}} \:\left(\mathrm{4}\:{edges}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\therefore{Number}\:{of}\:{rectangles}\:{with}\:{left}- \\ $$$$\:\:\:\:\:\:\:\:\:{upper}\:{corner}\:\mathrm{A}=\mathrm{2}×\mathrm{4}=\mathrm{8} \\ $$$$\:\:\:\:\:\:\:\:\:\left({Recall}\:{that}\:{every}\:{upper}\:{edge}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:{makes}\:{a}\:{rectangle}\:{with}\:{every} \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:{left}\:{edge}.\right) \\ $$$$\:\:\:\:\:\:\:\:\:....... \\ $$

Commented by Tony Lin last updated on 25/Aug/19

A:(AB,AC)×(AG,AL,AQ,AA_1 )=8 B:BC×(BI,BN,BS,BB_1 )=4 C:0 D:(DE,DF)×(DH,DM,DR,DV)=8 E:EF×(EI,EN,ES,EW)=4 F:0 G: { (((GI,GK)×(GL,GQ,GA_1 )=6)),(((GH,GJ)×(GL,GQ)=4)) :} H: { (((HI,HJ)×(HM,HR,HV)=6)),((HK×(HM,HR)=2)) :} I: { ((IJ×(IN,IS,IW)=3)),((IK×(IN,IS,IW)=3)) :} J:JK×(JO,JT)=2 K:0 L: { (((LN,LP)×(LQ,LA_1 )=4)),(((LM,LO)×LQ=2)) :} M: { (((MN,MO,MP)×MR=3)),(((MN,MO)×MV=2)) :} N: { ((NO×(NS,NW)=2)),((NP×(NS,NB_1 )=2)) :} O:OP×OT=1 P:0 Q:(QS,QU)×QA_1 =2 R:(RS,RT)×RV=2 S: { ((ST×SW=1)),((SU×SB_1 =1)) :} T∼C_1 :0 SUM:72 a little complex, but the result is the same this is the power of classification

$${A}:\left({AB},{AC}\right)×\left({AG},{AL},{AQ},{AA}_{\mathrm{1}} \right)=\mathrm{8} \\ $$$${B}:{BC}×\left({BI},{BN},{BS},{BB}_{\mathrm{1}} \right)=\mathrm{4} \\ $$$${C}:\mathrm{0} \\ $$$${D}:\left({DE},{DF}\right)×\left({DH},{DM},{DR},{DV}\right)=\mathrm{8} \\ $$$${E}:{EF}×\left({EI},{EN},{ES},{EW}\right)=\mathrm{4} \\ $$$${F}:\mathrm{0} \\ $$$${G}:\begin{cases}{\left({GI},{GK}\right)×\left({GL},{GQ},{GA}_{\mathrm{1}} \right)=\mathrm{6}}\\{\left({GH},{GJ}\right)×\left({GL},{GQ}\right)=\mathrm{4}}\end{cases} \\ $$$${H}:\begin{cases}{\left({HI},{HJ}\right)×\left({HM},{HR},{HV}\right)=\mathrm{6}}\\{{HK}×\left({HM},{HR}\right)=\mathrm{2}}\end{cases} \\ $$$${I}:\begin{cases}{{IJ}×\left({IN},{IS},{IW}\right)=\mathrm{3}}\\{{IK}×\left({IN},{IS},{IW}\right)=\mathrm{3}}\end{cases} \\ $$$${J}:{JK}×\left({JO},{JT}\right)=\mathrm{2} \\ $$$${K}:\mathrm{0} \\ $$$${L}:\begin{cases}{\left({LN},{LP}\right)×\left({LQ},{LA}_{\mathrm{1}} \right)=\mathrm{4}}\\{\left({LM},{LO}\right)×{LQ}=\mathrm{2}}\end{cases} \\ $$$${M}:\begin{cases}{\left({MN},{MO},{MP}\right)×{MR}=\mathrm{3}}\\{\left({MN},{MO}\right)×{MV}=\mathrm{2}}\end{cases} \\ $$$${N}:\begin{cases}{{NO}×\left({NS},{NW}\right)=\mathrm{2}}\\{{NP}×\left({NS},{NB}_{\mathrm{1}} \right)=\mathrm{2}}\end{cases} \\ $$$${O}:{OP}×{OT}=\mathrm{1} \\ $$$${P}:\mathrm{0} \\ $$$${Q}:\left({QS},{QU}\right)×{QA}_{\mathrm{1}} =\mathrm{2} \\ $$$${R}:\left({RS},{RT}\right)×{RV}=\mathrm{2} \\ $$$${S}:\begin{cases}{{ST}×{SW}=\mathrm{1}}\\{{SU}×{SB}_{\mathrm{1}} =\mathrm{1}}\end{cases} \\ $$$${T}\sim{C}_{\mathrm{1}} :\mathrm{0} \\ $$$${SUM}:\mathrm{72} \\ $$$${a}\:{little}\:{complex},\:{but}\:{the}\:{result}\:{is}\:{the}\:{same} \\ $$$${this}\:{is}\:{the}\:{power}\:{of}\:{classification} \\ $$

Commented by peter frank last updated on 25/Aug/19

thank you both

$${thank}\:{you}\:{both} \\ $$

Commented by Rasheed.Sindhi last updated on 25/Aug/19

Great Thanks for Great Help S∥^(⧫) R! I found my mistake after comparing my work with yours! Also thank ⋓ Peter Frank!

$$\mathcal{G}{reat}\:\mathcal{T}{hanks}\:{for}\:\mathcal{G}{reat}\:\mathcal{H}{elp}\:\mathrm{S}\overset{\blacklozenge} {\parallel}\mathrm{R}! \\ $$$$\mathcal{I}\:{found}\:{my}\:{mistake}\:{after}\:{comparing} \\ $$$${my}\:{work}\:{with}\:{yours}! \\ $$$$ \\ $$$${Also}\:{thank}\:\Cup\:{Peter}\:{Frank}! \\ $$

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