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Question Number 155109 by qaz last updated on 25/Sep/21
how many terms contain “ ab^2 c^3 ” in (a+2b+3c+4d^2 +5e^3 )^(10) ?
$$\mathrm{how}\:\mathrm{many}\:\mathrm{terms}\:\mathrm{contain}\:“\:\mathrm{ab}^{\mathrm{2}} \mathrm{c}^{\mathrm{3}} \:''\:\mathrm{in}\:\left(\mathrm{a}+\mathrm{2b}+\mathrm{3c}+\mathrm{4d}^{\mathrm{2}} +\mathrm{5e}^{\mathrm{3}} \right)^{\mathrm{10}} \:? \\ $$
Answered by mr W last updated on 26/Sep/21
(a+2b+3c+4d^2 +5e^3 )^(10) =Σ_(0≤α,β,γ,δ,ε≤10) ^(α+β+γ+δ+ε=10) ((( 10)),((α,β,γ,δ,ε)) )a^α (2b)^β (3c)^γ (4d^2 )^δ (5e^3 )^ε terms containing ab^2 c^3 fulfill α+β+γ+δ+ε=10 ...(i) with α≥1 ⇒x+x^2 +x^3 +...=x(1+x+x^2 +...) β≥2 ⇒x^2 +x^3 +x^4 ...=x^2 (1+x+x^2 +...) γ≥3 ⇒x^3 +x^4 +x^5 +...=x^3 (1+x+x^2 +...) δ≥0 ⇒1+x+x^2 +x^3 +...=(1+x+x^2 +...) ε≥0 ⇒1+x+x^2 +x^3 +...=(1+x+x^2 +...) x^6 (1+x+x^2 +...)^5 =(x^6 /((1−x)^5 ))=Σ_(k=0) ^∞ C_4 ^(k+4) x^(k+6) coef. of x^(10) is C_4 ^(4+4) =C_4 ^8 =70 that means (i) has 70 solutions satisfying given conditions. i.e. there are 70 terms containing ab^2 c^3 .
$$\left({a}+\mathrm{2}{b}+\mathrm{3}{c}+\mathrm{4}{d}^{\mathrm{2}} +\mathrm{5}{e}^{\mathrm{3}} \right)^{\mathrm{10}} \\ $$$$=\underset{\mathrm{0}\leqslant\alpha,\beta,\gamma,\delta,\epsilon\leqslant\mathrm{10}} {\overset{\alpha+\beta+\gamma+\delta+\epsilon=\mathrm{10}} {\sum}}\begin{pmatrix}{\:\:\:\:\:\:\:\:\mathrm{10}}\\{\alpha,\beta,\gamma,\delta,\epsilon}\end{pmatrix}{a}^{\alpha} \left(\mathrm{2}{b}\right)^{\beta} \left(\mathrm{3}{c}\right)^{\gamma} \left(\mathrm{4}{d}^{\mathrm{2}} \right)^{\delta} \left(\mathrm{5}{e}^{\mathrm{3}} \right)^{\epsilon} \\ $$$${terms}\:{containing}\:{ab}^{\mathrm{2}} {c}^{\mathrm{3}} \:{fulfill} \\ $$$$\alpha+\beta+\gamma+\delta+\epsilon=\mathrm{10}\:\:\:…\left({i}\right) \\ $$$${with} \\ $$$$\alpha\geqslant\mathrm{1}\:\Rightarrow{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…={x}\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…\right) \\ $$$$\beta\geqslant\mathrm{2}\:\Rightarrow{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} …={x}^{\mathrm{2}} \left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…\right) \\ $$$$\gamma\geqslant\mathrm{3}\:\Rightarrow{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +{x}^{\mathrm{5}} +…={x}^{\mathrm{3}} \left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…\right) \\ $$$$\delta\geqslant\mathrm{0}\:\Rightarrow\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…=\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…\right) \\ $$$$\epsilon\geqslant\mathrm{0}\:\Rightarrow\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…=\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…\right) \\ $$$$ \\ $$$${x}^{\mathrm{6}} \left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…\right)^{\mathrm{5}} =\frac{{x}^{\mathrm{6}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{5}} }=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{4}} ^{{k}+\mathrm{4}} {x}^{{k}+\mathrm{6}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{10}} \:{is}\:{C}_{\mathrm{4}} ^{\mathrm{4}+\mathrm{4}} ={C}_{\mathrm{4}} ^{\mathrm{8}} =\mathrm{70} \\ $$$${that}\:{means}\:\left({i}\right)\:{has}\:\mathrm{70}\:{solutions} \\ $$$${satisfying}\:{given}\:{conditions}.\:{i}.{e}.\: \\ $$$${there}\:{are}\:\mathrm{70}\:{terms}\:{containing}\:{ab}^{\mathrm{2}} {c}^{\mathrm{3}} . \\ $$
Commented by Tawa11 last updated on 26/Sep/21
great sir
$$\mathrm{great}\:\mathrm{sir} \\ $$

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