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Question Number 155109 by qaz last updated on 25/Sep/21

$$\mathrm{how}\mathrm{many}\mathrm{terms}\mathrm{contain}‘‘{\mathrm{ab}}^2{\mathrm{c}}^3{}^{″}\mathrm{in}{(\mathrm{a}+2\mathrm{b}+3\mathrm{c}+{4\mathrm{d}}^2+{5\mathrm{e}}^3)}^{10}?$$

Answered by mr W last updated on 26/Sep/21

$${(a+2b+3c+4d^2+5e^3)}^{10}$$$$=\underset{0⩽\alpha ,\beta ,\gamma ,\delta ,ϵ⩽10}{\sum ^{\alpha +\beta +\gamma +\delta +ϵ=10}}\left(\begin{array}{c}10\\ \alpha ,\beta ,\gamma ,\delta ,ϵ\end{array}\right)a^{\alpha }{\left(2b\right)}^{\beta }{\left(3c\right)}^{\gamma }{\left(4d^2\right)}^{\delta }{\left(5e^3\right)}^{ϵ}$$$$termscontaining{ab}^2{c}^{3}fulfill$$$$\alpha +\beta +\gamma +\delta +ϵ=10...\left(i\right)$$$$with$$$$\alpha ⩾1\Rightarrow x+x^2+x^3+...=x(1+x+x^2+...)$$$$\beta ⩾2\Rightarrow x^2+x^3+x^4...=x^2(1+x+x^2+...)$$$$\gamma ⩾3\Rightarrow x^3+x^4+x^5+...=x^3(1+x+x^2+...)$$$$\delta ⩾0\Rightarrow 1+x+x^2+x^3+...=(1+x+x^2+...)$$$$ϵ⩾0\Rightarrow 1+x+x^2+x^3+...=(1+x+x^2+...)$$$$$$$$x^6{(1+x+x^2+...)}^5=\frac{x^6}{{(1-x)}^5}=\underset{k=0}{\sum ^{\mathrm{\infty }}}{C}_4^{k+4}{x}^{k+6}$$$$coef.of{x}^{10}is{C}_4^{4+4}=C_4^8=70$$$$thatmeans\left(i\right)has70solutions$$$$satisfyinggivenconditions.i.e.$$$$thereare70termscontaining{ab}^2{c}^3.$$

Commented by Tawa11 last updated on 26/Sep/21

$$\mathrm{great}\mathrm{sir}$$

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