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Question Number 8217 by tawakalitu last updated on 02/Oct/16
what is the coefficient of x^3 in the expansion of (1 + x + x^2 + x^3 + x^4 + x^5 )^6
$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{x}^{\mathrm{3}} \:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion} \\ $$$$\mathrm{of}\:\left(\mathrm{1}\:+\:\mathrm{x}\:+\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{x}^{\mathrm{4}} \:+\:\mathrm{x}^{\mathrm{5}} \right)^{\mathrm{6}} \\ $$
Commented by 123456 last updated on 04/Oct/16
1∙2∙3∙4∙5∙6 1+1+1=1+2=3 3:6 1+2:6∙5=30 1+1+1:((6∙5∙4)/(3!))=((6∙5∙4)/(3∙2))=5∙4=20 6+30+20=56
$$\mathrm{1}\centerdot\mathrm{2}\centerdot\mathrm{3}\centerdot\mathrm{4}\centerdot\mathrm{5}\centerdot\mathrm{6} \\ $$$$\mathrm{1}+\mathrm{1}+\mathrm{1}=\mathrm{1}+\mathrm{2}=\mathrm{3} \\ $$$$\mathrm{3}:\mathrm{6} \\ $$$$\mathrm{1}+\mathrm{2}:\mathrm{6}\centerdot\mathrm{5}=\mathrm{30} \\ $$$$\mathrm{1}+\mathrm{1}+\mathrm{1}:\frac{\mathrm{6}\centerdot\mathrm{5}\centerdot\mathrm{4}}{\mathrm{3}!}=\frac{\mathrm{6}\centerdot\mathrm{5}\centerdot\mathrm{4}}{\mathrm{3}\centerdot\mathrm{2}}=\mathrm{5}\centerdot\mathrm{4}=\mathrm{20} \\ $$$$\mathrm{6}+\mathrm{30}+\mathrm{20}=\mathrm{56} \\ $$
Commented by sandy_suhendra last updated on 04/Oct/16
this is so simple, but can you more explain it?
$$\mathrm{this}\:\mathrm{is}\:\mathrm{so}\:\mathrm{simple},\:\mathrm{but}\:\mathrm{can}\:\mathrm{you}\:\mathrm{more}\:\mathrm{explain}\:\mathrm{it}? \\ $$
Commented by 123456 last updated on 04/Oct/16
we have (1+x+x^2 +x^3 +x^4 +x^5 )^6 = (1+∙∙∙+x^5 )...(1+∙∙∙+x^5 ) recal x^3 =x^2 x=xxx we have to pick one factor from each (), if we pick x^3 , all other will be 1 we have 6 () so 6 ways if we pick x^2 , we need to pick a x and all other will be 1, so 6() × 5() = 30 ways if we pick x, we pick x^2 (above case) or x and x, so 6 () × 5() × 4() =“120 ways” however x×x×x=x×x×x (3!=6 ways of perm x) so ((120)/6)=20 ways since all coef are 1 the coef is the sum of number of way 6+30+20=56
$$\mathrm{we}\:\mathrm{have} \\ $$$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +{x}^{\mathrm{5}} \right)^{\mathrm{6}} = \\ $$$$\left(\mathrm{1}+\centerdot\centerdot\centerdot+{x}^{\mathrm{5}} \right)…\left(\mathrm{1}+\centerdot\centerdot\centerdot+{x}^{\mathrm{5}} \right) \\ $$$$\mathrm{recal}\:{x}^{\mathrm{3}} ={x}^{\mathrm{2}} {x}={xxx} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{pick}\:\mathrm{one}\:\mathrm{factor}\:\mathrm{from}\:\mathrm{each} \\ $$$$\left(\right),\:\mathrm{if}\:\mathrm{we}\:\mathrm{pick}\:{x}^{\mathrm{3}} ,\:\mathrm{all}\:\mathrm{other}\:\mathrm{will}\:\mathrm{be}\:\mathrm{1} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{6}\:\left(\right)\:\mathrm{so}\:\mathrm{6}\:\mathrm{ways} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{pick}\:{x}^{\mathrm{2}} ,\:\mathrm{we}\:\mathrm{need}\:\mathrm{to}\:\mathrm{pick}\:\mathrm{a}\:{x}\:\mathrm{and} \\ $$$$\mathrm{all}\:\mathrm{other}\:\mathrm{will}\:\mathrm{be}\:\mathrm{1},\:\mathrm{so} \\ $$$$\mathrm{6}\left(\right)\:×\:\mathrm{5}\left(\right)\:=\:\mathrm{30}\:\mathrm{ways} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{pick}\:{x},\:\mathrm{we}\:\mathrm{pick}\:{x}^{\mathrm{2}} \:\left(\mathrm{above}\:\mathrm{case}\right) \\ $$$$\mathrm{or}\:{x}\:\mathrm{and}\:{x},\:\mathrm{so} \\ $$$$\mathrm{6}\:\left(\right)\:×\:\mathrm{5}\left(\right)\:×\:\mathrm{4}\left(\right)\:=“\mathrm{120}\:\mathrm{ways}'' \\ $$$$\mathrm{however} \\ $$$${x}×{x}×{x}={x}×{x}×{x}\:\left(\mathrm{3}!=\mathrm{6}\:\mathrm{ways}\:\mathrm{of}\:\mathrm{perm}\:{x}\right) \\ $$$$\mathrm{so} \\ $$$$\frac{\mathrm{120}}{\mathrm{6}}=\mathrm{20}\:\mathrm{ways} \\ $$$$\mathrm{since}\:\mathrm{all}\:\mathrm{coef}\:\mathrm{are}\:\mathrm{1}\:\mathrm{the}\:\mathrm{coef}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum} \\ $$$$\mathrm{of}\:\mathrm{number}\:\mathrm{of}\:\mathrm{way} \\ $$$$\mathrm{6}+\mathrm{30}+\mathrm{20}=\mathrm{56} \\ $$
Commented by sandy_suhendra last updated on 04/Oct/16
great! thank′s for your answer
$$\mathrm{great}!\:\mathrm{thank}'\mathrm{s}\:\mathrm{for}\:\mathrm{your}\:\mathrm{answer} \\ $$
Commented by tawakalitu last updated on 04/Oct/16
Thanks so much. i really appreciate.
$$\mathrm{Thanks}\:\mathrm{so}\:\mathrm{much}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$
Answered by sandy_suhendra last updated on 03/Oct/16
[(1+x+x^2 )+x^3 (1+x+x^2 )]^6 =[(x^3 +1)(x^2 +x+1)]^6 =(x^3 +1)^6 (x^2 +x+1)^6 (x^3 +1)^6 =(x^3 )^6 +6(x^3 )^5 +...+6x^3 +1 [x^2 +(x+1)]^6 =1(x^2 )^6 +6(x^2 )^5 (x+1)+...+6x^2 (x+1)^5 +1(x+1)^6 6x^2 (x+1)^5 =6x^2 (x^5 +5x^4 +...+5x+1)⇒30x^3 (x+1)^6 =x^6 +6x^5 +...+20x^3 +...+1 the term of x^3 comes from ⇒ 6x^3 +30x^3 +20x^3 =56x^3 so the coefficient of x^3 is 56
$$\left[\left(\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} \right)+\mathrm{x}^{\mathrm{3}} \left(\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} \right)\right]^{\mathrm{6}} \\ $$$$=\left[\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)\right]^{\mathrm{6}} \\ $$$$=\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{6}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)^{\mathrm{6}} \\ $$$$\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{6}} =\left(\mathrm{x}^{\mathrm{3}} \right)^{\mathrm{6}} +\mathrm{6}\left(\mathrm{x}^{\mathrm{3}} \right)^{\mathrm{5}} +…+\mathrm{6x}^{\mathrm{3}} +\mathrm{1} \\ $$$$\left[\mathrm{x}^{\mathrm{2}} +\left(\mathrm{x}+\mathrm{1}\right)\right]^{\mathrm{6}} \\ $$$$=\mathrm{1}\left(\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{6}} +\mathrm{6}\left(\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{5}} \left(\mathrm{x}+\mathrm{1}\right)+…+\mathrm{6x}^{\mathrm{2}} \left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{5}} +\mathrm{1}\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{6}} \\ $$$$\mathrm{6x}^{\mathrm{2}} \left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{5}} =\mathrm{6x}^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{5}} +\mathrm{5x}^{\mathrm{4}} +…+\mathrm{5}\boldsymbol{\mathrm{x}}+\mathrm{1}\right)\Rightarrow\mathrm{30x}^{\mathrm{3}} \\ $$$$\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{6}} =\mathrm{x}^{\mathrm{6}} +\mathrm{6x}^{\mathrm{5}} +…+\mathrm{20x}^{\mathrm{3}} +…+\mathrm{1} \\ $$$$\mathrm{the}\:\mathrm{term}\:\mathrm{of}\:\mathrm{x}^{\mathrm{3}} \:\mathrm{comes}\:\mathrm{from}\:\Rightarrow\:\mathrm{6x}^{\mathrm{3}} +\mathrm{30x}^{\mathrm{3}} +\mathrm{20x}^{\mathrm{3}} =\mathrm{56x}^{\mathrm{3}} \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{x}^{\mathrm{3}} \:\mathrm{is}\:\mathrm{56} \\ $$
Commented by tawakalitu last updated on 03/Oct/16
Thanks so much.
$$\mathrm{Thanks}\:\mathrm{so}\:\mathrm{much}. \\ $$

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