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Question Number 8217 by tawakalitu last updated on 02/Oct/16

$$\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^3\mathrm{in}\mathrm{the}\mathrm{expansion}$$$$\mathrm{of}{(1+\mathrm{x}+{\mathrm{x}}^2+{\mathrm{x}}^3+{\mathrm{x}}^4+{\mathrm{x}}^5)}^6$$

Commented by 123456 last updated on 04/Oct/16

$$1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6$$$$1+1+1=1+2=3$$$$3:6$$$$1+2:6\cdot 5=30$$$$1+1+1:\frac{6\cdot 5\cdot 4}{3!}=\frac{6\cdot 5\cdot 4}{3\cdot 2}=5\cdot 4=20$$$$6+30+20=56$$

Commented by sandy_suhendra last updated on 04/Oct/16

$$\mathrm{this}\mathrm{is}\mathrm{so}\mathrm{simple},\mathrm{but}\mathrm{can}\mathrm{you}\mathrm{more}\mathrm{explain}\mathrm{it}?$$

Commented by 123456 last updated on 04/Oct/16

$$\mathrm{we}\mathrm{have}$$$${(1+x+x^2+x^3+x^4+x^5)}^6=$$$$(1+\cdot \cdot \cdot +x^5)...(1+\cdot \cdot \cdot +x^5)$$$$\mathrm{recal}{x}^3=x^{2}x=xxx$$$$\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{pick}\mathrm{one}\mathrm{factor}\mathrm{from}\mathrm{each}$$$$\left(\right),\mathrm{if}\mathrm{we}\mathrm{pick}{x}^3,\mathrm{all}\mathrm{other}\mathrm{will}\mathrm{be}1$$$$\mathrm{we}\mathrm{have}6\left(\right)\mathrm{so}6\mathrm{ways}$$$$\mathrm{if}\mathrm{we}\mathrm{pick}{x}^2,\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{pick}\mathrm{a}x\mathrm{and}$$$$\mathrm{all}\mathrm{other}\mathrm{will}\mathrm{be}1,\mathrm{so}$$$$6\left(\right)\times 5\left(\right)=30\mathrm{ways}$$$$\mathrm{if}\mathrm{we}\mathrm{pick}x,\mathrm{we}\mathrm{pick}{x}^2\left(\mathrm{above}\mathrm{case}\right)$$$$\mathrm{or}x\mathrm{and}x,\mathrm{so}$$$$6\left(\right)\times 5\left(\right)\times 4\left(\right)=‘‘120{\mathrm{ways}}^{″}$$$$\mathrm{however}$$$$x\times x\times x=x\times x\times x(3!=6\mathrm{ways}\mathrm{of}\mathrm{perm}x)$$$$\mathrm{so}$$$$\frac{120}{6}=20\mathrm{ways}$$$$\mathrm{since}\mathrm{all}\mathrm{coef}\mathrm{are}1\mathrm{the}\mathrm{coef}\mathrm{is}\mathrm{the}\mathrm{sum}$$$$\mathrm{of}\mathrm{number}\mathrm{of}\mathrm{way}$$$$6+30+20=56$$

Commented by sandy_suhendra last updated on 04/Oct/16

$$\mathrm{great}!{\mathrm{thank}}^{\prime }\mathrm{s}\mathrm{for}\mathrm{your}\mathrm{answer}$$

Commented by tawakalitu last updated on 04/Oct/16

$$\mathrm{Thanks}\mathrm{so}\mathrm{much}.\mathrm{i}\mathrm{really}\mathrm{appreciate}.$$

Answered by sandy_suhendra last updated on 03/Oct/16

$${[(1+\mathrm{x}+{\mathrm{x}}^2)+{\mathrm{x}}^3(1+\mathrm{x}+{\mathrm{x}}^2)]}^6$$$$={\left[({\mathrm{x}}^3+1)({\mathrm{x}}^2+\mathrm{x}+1)\right]}^6$$$$={({\mathrm{x}}^3+1)}^6{({\mathrm{x}}^2+\mathrm{x}+1)}^6$$$${({\mathrm{x}}^3+1)}^6={\left({\mathrm{x}}^3\right)}^6+6{\left({\mathrm{x}}^3\right)}^5+...+{6\mathrm{x}}^3+1$$$${[{\mathrm{x}}^2+(\mathrm{x}+1)]}^6$$$$=1{\left({\mathrm{x}}^2\right)}^6+6{\left({\mathrm{x}}^2\right)}^5(\mathrm{x}+1)+...+{6\mathrm{x}}^2{(\mathrm{x}+1)}^5+1{(\mathrm{x}+1)}^6$$$${6\mathrm{x}}^2{(\mathrm{x}+1)}^5={6\mathrm{x}}^2({\mathrm{x}}^5+{5\mathrm{x}}^4+...+5\mathbf{x}+1)\Rightarrow {30\mathrm{x}}^3$$$${(\mathrm{x}+1)}^6={\mathrm{x}}^6+{6\mathrm{x}}^5+...+{20\mathrm{x}}^3+...+1$$$$\mathrm{the}\mathrm{term}\mathrm{of}{\mathrm{x}}^3\mathrm{comes}\mathrm{from}\Rightarrow {6\mathrm{x}}^3+{30\mathrm{x}}^3+{20\mathrm{x}}^3={56\mathrm{x}}^3$$$$\mathrm{so}\mathrm{the}\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^3\mathrm{is}56$$

Commented by tawakalitu last updated on 03/Oct/16

$$\mathrm{Thanks}\mathrm{so}\mathrm{much}.$$

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