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If-a-2-a-C-2-a-2-a-C-4-then-a-




Question Number 97737 by bagjamath last updated on 09/Jun/20
If ^((a^2 −a)) C_2 =^((a^2 −a)) C_4  , then a =
$$\mathrm{If}\:\:^{\left({a}^{\mathrm{2}} −{a}\right)} {C}_{\mathrm{2}} =\:^{\left({a}^{\mathrm{2}} −{a}\right)} {C}_{\mathrm{4}} \:,\:\mathrm{then}\:{a}\:= \\ $$
Commented by som(math1967) last updated on 09/Jun/20
a=3
$$\mathrm{a}=\mathrm{3} \\ $$
Commented by smridha last updated on 09/Jun/20
you have to put..in which set  does ′a′ belong??
$${you}\:\boldsymbol{{have}}\:\boldsymbol{{to}}\:\boldsymbol{{put}}..\boldsymbol{{in}}\:\boldsymbol{{which}}\:\boldsymbol{{set}} \\ $$$$\boldsymbol{{does}}\:'\boldsymbol{{a}}'\:\boldsymbol{{belong}}?? \\ $$
Commented by mr W last updated on 09/Jun/20
a^2 −a=2+4=6  (a+2)(a−3)=0  ⇒a=3
$${a}^{\mathrm{2}} −{a}=\mathrm{2}+\mathrm{4}=\mathrm{6} \\ $$$$\left({a}+\mathrm{2}\right)\left({a}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow{a}=\mathrm{3} \\ $$
Answered by smridha last updated on 09/Jun/20
(((a^2 −a)!)/(2!(a^2 −a−2)!))=(((a^2 −a)!)/(4!(a^2 −a−4)!))  (a^2 −a−2)(a^2 −a−3)=12  let (a^2 −a−2)=x  x^2 −x−12=0 x=((1+_− 7)/2) x=(4,−3)  for x=4  a^2 −a−6=0 a=((1+_− 5)/2) so a=(3,−2)  for x=−3  a^2 −a+1=0  so a=((1+_− i(√3))/2) a=(e^((i𝛑)/3) ,e^((−i𝛑)/3) )
$$\frac{\left(\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{a}}\right)!}{\mathrm{2}!\left(\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{a}}−\mathrm{2}\right)!}=\frac{\left({a}^{\mathrm{2}} −\boldsymbol{{a}}\right)!}{\mathrm{4}!\left(\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{a}}−\mathrm{4}\right)!} \\ $$$$\left(\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{a}}−\mathrm{2}\right)\left(\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{a}}−\mathrm{3}\right)=\mathrm{12} \\ $$$$\boldsymbol{{let}}\:\left(\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{a}}−\mathrm{2}\right)=\boldsymbol{{x}} \\ $$$$\boldsymbol{{x}}^{\mathrm{2}} −\boldsymbol{{x}}−\mathrm{12}=\mathrm{0}\:\boldsymbol{{x}}=\frac{\mathrm{1}\underset{−} {+}\mathrm{7}}{\mathrm{2}}\:\boldsymbol{{x}}=\left(\mathrm{4},−\mathrm{3}\right) \\ $$$$\boldsymbol{{for}}\:\boldsymbol{{x}}=\mathrm{4} \\ $$$$\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{a}}−\mathrm{6}=\mathrm{0}\:\boldsymbol{{a}}=\frac{\mathrm{1}\underset{−} {+}\mathrm{5}}{\mathrm{2}}\:\boldsymbol{{so}}\:\boldsymbol{{a}}=\left(\mathrm{3},−\mathrm{2}\right) \\ $$$${for}\:\boldsymbol{{x}}=−\mathrm{3} \\ $$$$\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{a}}+\mathrm{1}=\mathrm{0}\:\:\boldsymbol{{so}}\:\boldsymbol{{a}}=\frac{\mathrm{1}\underset{−} {+}\boldsymbol{{i}}\sqrt{\mathrm{3}}}{\mathrm{2}}\:\boldsymbol{{a}}=\left(\boldsymbol{{e}}^{\frac{\boldsymbol{{i}\pi}}{\mathrm{3}}} ,\boldsymbol{{e}}^{\frac{−\boldsymbol{{i}\pi}}{\mathrm{3}}} \right) \\ $$$$ \\ $$
Commented by smridha last updated on 09/Jun/20
I am not fighting with others...  but you tried to do so I think..  I just wanted to tell two simple  things (i)I am not claim all the  possible values of  a are  the   solutions(ii) even I didnot conclude  anything so why you are the guys  are so worrid with that...  am I criticize anyone??if this  is so simple matter then it is not  requare furthur comment.
$$\boldsymbol{{I}}\:{am}\:\boldsymbol{{not}}\:\boldsymbol{{fighting}}\:\boldsymbol{{with}}\:\boldsymbol{{others}}… \\ $$$$\boldsymbol{{but}}\:\boldsymbol{{you}}\:\boldsymbol{{tried}}\:\boldsymbol{{to}}\:\boldsymbol{{do}}\:\boldsymbol{{so}}\:\boldsymbol{{I}}\:{think}.. \\ $$$$\boldsymbol{{I}}\:{just}\:\boldsymbol{{wanted}}\:\boldsymbol{{to}}\:\boldsymbol{{tell}}\:\boldsymbol{{two}}\:\boldsymbol{{simple}} \\ $$$$\boldsymbol{{things}}\:\left(\boldsymbol{{i}}\right)\boldsymbol{{I}}\:{am}\:\boldsymbol{{not}}\:\boldsymbol{{claim}}\:\boldsymbol{{all}}\:\boldsymbol{{the}} \\ $$$$\boldsymbol{{possible}}\:\boldsymbol{{values}}\:\boldsymbol{{of}}\:\:\boldsymbol{{a}}\:\boldsymbol{{are}}\:\:\boldsymbol{{the}}\: \\ $$$$\boldsymbol{{solutions}}\left(\boldsymbol{{ii}}\right)\:\boldsymbol{{even}}\:\boldsymbol{{I}}\:{d}\boldsymbol{{id}}{not}\:{con}\boldsymbol{{clude}} \\ $$$$\boldsymbol{{anything}}\:\boldsymbol{{so}}\:\boldsymbol{{why}}\:\boldsymbol{{you}}\:\boldsymbol{{are}}\:\boldsymbol{{the}}\:\boldsymbol{{guys}} \\ $$$$\boldsymbol{{are}}\:\boldsymbol{{so}}\:\boldsymbol{{worrid}}\:\boldsymbol{{with}}\:\boldsymbol{{that}}… \\ $$$$\boldsymbol{{am}}\:\boldsymbol{{I}}\:{criti}\boldsymbol{{cize}}\:\boldsymbol{{anyone}}??\boldsymbol{{if}}\:\boldsymbol{{this}} \\ $$$$\boldsymbol{{is}}\:\boldsymbol{{so}}\:\boldsymbol{{simple}}\:\boldsymbol{{matter}}\:\boldsymbol{{then}}\:\boldsymbol{{it}}\:\boldsymbol{{is}}\:\boldsymbol{{not}} \\ $$$$\boldsymbol{{requare}}\:\boldsymbol{{furthur}}\:\boldsymbol{{comment}}. \\ $$
Commented by mathmax by abdo last updated on 09/Jun/20
how to get complex number in combination...!
$$\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\mathrm{complex}\:\mathrm{number}\:\mathrm{in}\:\mathrm{combination}…! \\ $$
Commented by smridha last updated on 09/Jun/20
am I put complex number in  combination???then what??  I just show how many values  a have??I don′t claim all the   possible values must satisfy  the combination......  the apperence of this question is  wrong!!! find a???what does it   mean??we logically choose  the value of a among the values  of a.so it does not mean that a  have only possible value 3..   moreover it means a takes value  3 for satisfy this combination.  for this combination 3 is unique  value  (of a) but not for a.
$$\boldsymbol{{am}}\:\boldsymbol{{I}}\:\boldsymbol{{put}}\:\boldsymbol{{complex}}\:\boldsymbol{{number}}\:\boldsymbol{{in}} \\ $$$$\boldsymbol{{combination}}???\boldsymbol{{then}}\:\boldsymbol{{what}}?? \\ $$$$\boldsymbol{{I}}\:\boldsymbol{{just}}\:\boldsymbol{{show}}\:\boldsymbol{{how}}\:\boldsymbol{{many}}\:\boldsymbol{{values}} \\ $$$$\boldsymbol{{a}}\:\boldsymbol{{have}}??\boldsymbol{{I}}\:\boldsymbol{{don}}'\boldsymbol{{t}}\:\boldsymbol{{claim}}\:\boldsymbol{{all}}\:\boldsymbol{{the}}\: \\ $$$$\boldsymbol{{possible}}\:\boldsymbol{{values}}\:\boldsymbol{{must}}\:\boldsymbol{{satisfy}} \\ $$$$\boldsymbol{{the}}\:\boldsymbol{{combination}}…… \\ $$$$\boldsymbol{{the}}\:\boldsymbol{{apperence}}\:\boldsymbol{{of}}\:\boldsymbol{{this}}\:\boldsymbol{{question}}\:\boldsymbol{{is}} \\ $$$$\boldsymbol{{wrong}}!!!\:\boldsymbol{{find}}\:\boldsymbol{{a}}???\boldsymbol{{what}}\:\boldsymbol{{does}}\:\boldsymbol{{it}}\: \\ $$$$\boldsymbol{{mean}}??\boldsymbol{{we}}\:\boldsymbol{{log}}{i}\boldsymbol{{cally}}\:\boldsymbol{{choose}} \\ $$$$\boldsymbol{{the}}\:\boldsymbol{{value}}\:\boldsymbol{{of}}\:\boldsymbol{{a}}\:\boldsymbol{{among}}\:\boldsymbol{{the}}\:\boldsymbol{{values}} \\ $$$$\boldsymbol{{of}}\:\boldsymbol{{a}}.\boldsymbol{{so}}\:\boldsymbol{{it}}\:\boldsymbol{{does}}\:\boldsymbol{{not}}\:\boldsymbol{{mean}}\:\boldsymbol{{that}}\:\boldsymbol{{a}} \\ $$$$\boldsymbol{{have}}\:\boldsymbol{{only}}\:\boldsymbol{{possible}}\:\boldsymbol{{value}}\:\mathrm{3}..\: \\ $$$$\boldsymbol{{moreover}}\:\boldsymbol{{it}}\:\boldsymbol{{means}}\:\boldsymbol{{a}}\:\boldsymbol{{takes}}\:\boldsymbol{{value}} \\ $$$$\mathrm{3}\:\boldsymbol{{for}}\:\boldsymbol{{satisfy}}\:\boldsymbol{{this}}\:\boldsymbol{{combination}}. \\ $$$$\boldsymbol{{for}}\:\boldsymbol{{this}}\:\boldsymbol{{combination}}\:\mathrm{3}\:\boldsymbol{{is}}\:\boldsymbol{{unique}} \\ $$$$\boldsymbol{{value}}\:\:\left(\boldsymbol{{of}}\:\boldsymbol{{a}}\right)\:\boldsymbol{{but}}\:\boldsymbol{{not}}\:\boldsymbol{{for}}\:\boldsymbol{{a}}. \\ $$
Commented by mr W last updated on 09/Jun/20
as for the answer of this question,  the only correct answer is a=3, other  values of a are non−sense for this  question.  i think as C_2 ^(a^2 −a) =C_4 ^(a^2 −a)  is written, it is  automatically said due to definition  that a∈Z and a^2 −a≥2 and a^2 −a≥4.
$${as}\:{for}\:{the}\:{answer}\:{of}\:{this}\:{question}, \\ $$$${the}\:{only}\:{correct}\:{answer}\:{is}\:{a}=\mathrm{3},\:{other} \\ $$$${values}\:{of}\:{a}\:{are}\:{non}−{sense}\:{for}\:{this} \\ $$$${question}. \\ $$$${i}\:{think}\:{as}\:{C}_{\mathrm{2}} ^{{a}^{\mathrm{2}} −{a}} ={C}_{\mathrm{4}} ^{{a}^{\mathrm{2}} −{a}} \:{is}\:{written},\:{it}\:{is} \\ $$$${automatically}\:{said}\:{due}\:{to}\:{definition} \\ $$$${that}\:\boldsymbol{{a}}\in\mathbb{Z}\:{and}\:{a}^{\mathrm{2}} −{a}\geqslant\mathrm{2}\:{and}\:\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{a}}\geqslant\mathrm{4}. \\ $$
Commented by smridha last updated on 09/Jun/20
ooh man!!am i claim all the possible  values satisfy this combination??  even am I put any conclution in my   solution??  this is the question of how much  we can do??  we have any idea??about  the factorial of complex number!!  factorial of negetive intiger!!  so we ignore them....this is the  fault.if there is theory behind   them we are then satisfied..is not  it????
$$\boldsymbol{{ooh}}\:\boldsymbol{{man}}!!\boldsymbol{{am}}\:\boldsymbol{{i}}\:\boldsymbol{{claim}}\:\boldsymbol{{all}}\:\boldsymbol{{the}}\:\boldsymbol{{possible}} \\ $$$$\boldsymbol{{values}}\:\boldsymbol{{satisfy}}\:\boldsymbol{{this}}\:\boldsymbol{{combination}}?? \\ $$$$\boldsymbol{{even}}\:\boldsymbol{{am}}\:\boldsymbol{{I}}\:\boldsymbol{{put}}\:\boldsymbol{{any}}\:\boldsymbol{{conclution}}\:\boldsymbol{{in}}\:\boldsymbol{{my}} \\ $$$$\:\boldsymbol{{solution}}?? \\ $$$$\boldsymbol{{this}}\:\boldsymbol{{is}}\:\boldsymbol{{the}}\:\boldsymbol{{question}}\:\boldsymbol{{of}}\:\boldsymbol{{how}}\:\boldsymbol{{much}} \\ $$$${we}\:\boldsymbol{{can}}\:\boldsymbol{{do}}?? \\ $$$$\boldsymbol{{we}}\:\boldsymbol{{have}}\:\boldsymbol{{any}}\:\boldsymbol{{idea}}??\boldsymbol{{about}} \\ $$$$\boldsymbol{{the}}\:\boldsymbol{{factorial}}\:\boldsymbol{{of}}\:\boldsymbol{{complex}}\:\boldsymbol{{number}}!! \\ $$$$\boldsymbol{{factorial}}\:\boldsymbol{{of}}\:\boldsymbol{{negetive}}\:\boldsymbol{{intiger}}!! \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{we}}\:\boldsymbol{{ignore}}\:\boldsymbol{{them}}….\boldsymbol{{this}}\:\boldsymbol{{is}}\:\boldsymbol{{the}} \\ $$$$\boldsymbol{{fault}}.\boldsymbol{{if}}\:\boldsymbol{{there}}\:\boldsymbol{{is}}\:\boldsymbol{{theory}}\:\boldsymbol{{behind}}\: \\ $$$$\boldsymbol{{them}}\:\boldsymbol{{we}}\:\boldsymbol{{are}}\:\boldsymbol{{then}}\:\boldsymbol{{satisfied}}..\boldsymbol{{is}}\:\boldsymbol{{not}} \\ $$$$\boldsymbol{{it}}???? \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 09/Jun/20
sir: nobody said that you are wrong  or something like that. it seems  you see critical comments from others  as attacks and want to fight back.  what a pity! since a forum is place for  discussion,exchange and learning  from each other. all people here are  equal.
$${sir}:\:{nobody}\:{said}\:{that}\:{you}\:{are}\:{wrong} \\ $$$${or}\:{something}\:{like}\:{that}.\:{it}\:{seems} \\ $$$${you}\:{see}\:{critical}\:{comments}\:{from}\:{others} \\ $$$${as}\:{attacks}\:{and}\:{want}\:{to}\:{fight}\:{back}. \\ $$$${what}\:{a}\:{pity}!\:{since}\:{a}\:{forum}\:{is}\:{place}\:{for} \\ $$$${discussion},{exchange}\:{and}\:{learning} \\ $$$${from}\:{each}\:{other}.\:{all}\:{people}\:{here}\:{are} \\ $$$${equal}. \\ $$
Commented by abdomathmax last updated on 09/Jun/20
your answer is not correct
$$\mathrm{your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{not}\:\mathrm{correct}\: \\ $$
Commented by smridha last updated on 09/Jun/20
Mr Abdomathmax :am I write any values as ans??  if you think that this is your   fult not mine..!!
$$\boldsymbol{{M}}{r}\:{Ab}\boldsymbol{{domathmax}}\::\boldsymbol{{am}}\:\boldsymbol{{I}}\:{write}\:\boldsymbol{{any}}\:\boldsymbol{{values}}\:\boldsymbol{{as}}\:\boldsymbol{{ans}}?? \\ $$$$\boldsymbol{{if}}\:\boldsymbol{{you}}\:\boldsymbol{{think}}\:\boldsymbol{{that}}\:\boldsymbol{{this}}\:\boldsymbol{{is}}\:\boldsymbol{{your}}\: \\ $$$$\boldsymbol{{fult}}\:\boldsymbol{{not}}\:\boldsymbol{{mine}}..!! \\ $$
Commented by mathmax by abdo last updated on 09/Jun/20
dont worry and be happy we are here to help  and learn...maths is a sea  without borders....!
$$\mathrm{dont}\:\mathrm{worry}\:\mathrm{and}\:\mathrm{be}\:\mathrm{happy}\:\mathrm{we}\:\mathrm{are}\:\mathrm{here}\:\mathrm{to}\:\mathrm{help}\:\:\mathrm{and}\:\mathrm{learn}…\mathrm{maths}\:\mathrm{is}\:\mathrm{a}\:\mathrm{sea} \\ $$$$\mathrm{without}\:\mathrm{borders}….! \\ $$
Answered by mathmax by abdo last updated on 09/Jun/20
C_(a^2 −a) ^2  =C_(a^2 −a) ^4  ⇒(((a^2 −a)!)/(2!(a^2 −a−2)!)) =(((a^2 −a)!)/(4!(a^2 −a−4)!))  (a≥2) ⇒  2!(a^2 −a−2)! =4! (a^2 −a−4)! ⇒  2(a^2 −a−2)(a^2 −a−3) =4! ⇒(a^2 −a−2)(a^2 −a−3) =((4.3.2)/2) =12  let a^2 −a−2 =2u and a^2 −a−3 =2v+1 ⇒2v +1 =2u−1 ⇒  2u.(2v+1) =12 ⇒u(2u−1) =6 ⇒2u^2 −u−6 =0  (u integr natural)  Δ =1−4×2×(−6) =1+48 =49 ⇒u_1 =((1+7)/4) =2  u_2 =((1−7)/4) =−(3/2)(to elminate) ⇒u=2 ⇒a^2 −a =2+4 =6 ⇒a^2 −a−6 =0  Δ=1−4(−6) =25 ⇒a =((1+5)/2) =3  or a =((1−5)/2) =−2<0(to eliminate)  the unique value is a =3
$$\mathrm{C}_{\mathrm{a}^{\mathrm{2}} −\mathrm{a}} ^{\mathrm{2}} \:=\mathrm{C}_{\mathrm{a}^{\mathrm{2}} −\mathrm{a}} ^{\mathrm{4}} \:\Rightarrow\frac{\left(\mathrm{a}^{\mathrm{2}} −\mathrm{a}\right)!}{\mathrm{2}!\left(\mathrm{a}^{\mathrm{2}} −\mathrm{a}−\mathrm{2}\right)!}\:=\frac{\left(\mathrm{a}^{\mathrm{2}} −\mathrm{a}\right)!}{\mathrm{4}!\left(\mathrm{a}^{\mathrm{2}} −\mathrm{a}−\mathrm{4}\right)!}\:\:\left(\mathrm{a}\geqslant\mathrm{2}\right)\:\Rightarrow \\ $$$$\mathrm{2}!\left(\mathrm{a}^{\mathrm{2}} −\mathrm{a}−\mathrm{2}\right)!\:=\mathrm{4}!\:\left(\mathrm{a}^{\mathrm{2}} −\mathrm{a}−\mathrm{4}\right)!\:\Rightarrow \\ $$$$\mathrm{2}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{a}−\mathrm{2}\right)\left(\mathrm{a}^{\mathrm{2}} −\mathrm{a}−\mathrm{3}\right)\:=\mathrm{4}!\:\Rightarrow\left(\mathrm{a}^{\mathrm{2}} −\mathrm{a}−\mathrm{2}\right)\left(\mathrm{a}^{\mathrm{2}} −\mathrm{a}−\mathrm{3}\right)\:=\frac{\mathrm{4}.\mathrm{3}.\mathrm{2}}{\mathrm{2}}\:=\mathrm{12} \\ $$$$\mathrm{let}\:\mathrm{a}^{\mathrm{2}} −\mathrm{a}−\mathrm{2}\:=\mathrm{2u}\:\mathrm{and}\:\mathrm{a}^{\mathrm{2}} −\mathrm{a}−\mathrm{3}\:=\mathrm{2v}+\mathrm{1}\:\Rightarrow\mathrm{2v}\:+\mathrm{1}\:=\mathrm{2u}−\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{2u}.\left(\mathrm{2v}+\mathrm{1}\right)\:=\mathrm{12}\:\Rightarrow\mathrm{u}\left(\mathrm{2u}−\mathrm{1}\right)\:=\mathrm{6}\:\Rightarrow\mathrm{2u}^{\mathrm{2}} −\mathrm{u}−\mathrm{6}\:=\mathrm{0}\:\:\left(\mathrm{u}\:\mathrm{integr}\:\mathrm{natural}\right) \\ $$$$\Delta\:=\mathrm{1}−\mathrm{4}×\mathrm{2}×\left(−\mathrm{6}\right)\:=\mathrm{1}+\mathrm{48}\:=\mathrm{49}\:\Rightarrow\mathrm{u}_{\mathrm{1}} =\frac{\mathrm{1}+\mathrm{7}}{\mathrm{4}}\:=\mathrm{2} \\ $$$$\mathrm{u}_{\mathrm{2}} =\frac{\mathrm{1}−\mathrm{7}}{\mathrm{4}}\:=−\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{to}\:\mathrm{elminate}\right)\:\Rightarrow\mathrm{u}=\mathrm{2}\:\Rightarrow\mathrm{a}^{\mathrm{2}} −\mathrm{a}\:=\mathrm{2}+\mathrm{4}\:=\mathrm{6}\:\Rightarrow\mathrm{a}^{\mathrm{2}} −\mathrm{a}−\mathrm{6}\:=\mathrm{0} \\ $$$$\Delta=\mathrm{1}−\mathrm{4}\left(−\mathrm{6}\right)\:=\mathrm{25}\:\Rightarrow\mathrm{a}\:=\frac{\mathrm{1}+\mathrm{5}}{\mathrm{2}}\:=\mathrm{3}\:\:\mathrm{or}\:\mathrm{a}\:=\frac{\mathrm{1}−\mathrm{5}}{\mathrm{2}}\:=−\mathrm{2}<\mathrm{0}\left(\mathrm{to}\:\mathrm{eliminate}\right) \\ $$$$\mathrm{the}\:\mathrm{unique}\:\mathrm{value}\:\mathrm{is}\:\mathrm{a}\:=\mathrm{3} \\ $$$$ \\ $$

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