Question Number 204815 by cortano12 last updated on 28/Feb/24
$$\:\:\:\:\:\mathrm{Given}\:\left(\mathrm{3p}^{\mathrm{2}} −\mathrm{p}+\mathrm{q}^{\mathrm{3}} \right)^{\mathrm{12}} \:,\:\mathrm{find}\:\mathrm{the}\: \\ $$$$\:\:\:\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{p}^{\mathrm{10}} \mathrm{q}^{\mathrm{6}} \\ $$
Answered by TonyCWX08 last updated on 28/Feb/24
$$ \\ $$$$\begin{array}{|c|c|c|c|c|}{\mathrm{3}{p}^{\mathrm{2}} }&\hline{−{p}}&\hline{{q}^{\mathrm{3}} }\\{\mathrm{4}}&\hline{\mathrm{2}}&\hline{\mathrm{2}}\\{\mathrm{3}}&\hline{\mathrm{4}}&\hline{\mathrm{2}}\\{\mathrm{2}}&\hline{\mathrm{6}}&\hline{\mathrm{2}}\\{\mathrm{1}}&\hline{\mathrm{8}}&\hline{\mathrm{2}}\\\hline\end{array}\: \\ $$$$ \\ $$$$\mathrm{12}+\mathrm{9}+\mathrm{6}+\mathrm{3} \\ $$$$=\mathrm{30} \\ $$
Answered by mr W last updated on 28/Feb/24
$$\left(\mathrm{3}{p}^{\mathrm{2}} −{p}+{q}^{\mathrm{3}} \right)^{\mathrm{12}} \\ $$$$=\underset{{a}+{b}+{c}=\mathrm{12}} {\sum}\begin{pmatrix}{\:\:\mathrm{12}}\\{{a},{b},{c}}\end{pmatrix}\left(\mathrm{3}{p}^{\mathrm{2}} \right)^{{a}} \left(−{p}\right)^{{b}} \left({q}^{\mathrm{3}} \right)^{{c}} \\ $$$$\mathrm{3}{c}=\mathrm{6}\:\Rightarrow\:{c}=\mathrm{2} \\ $$$${a}+{b}=\mathrm{12}−\mathrm{2}=\mathrm{10} \\ $$$$\mathrm{2}{a}+{b}=\mathrm{10} \\ $$$$\Rightarrow{a}=\mathrm{0},\:{b}=\mathrm{10} \\ $$$${coef}.\:{of}\:{p}^{\mathrm{10}} {q}^{\mathrm{6}} \:{is} \\ $$$$\begin{pmatrix}{\:\:\:\:\:\mathrm{12}}\\{\mathrm{0},\:\mathrm{10},\:\mathrm{2}}\end{pmatrix}\:\mathrm{3}^{\mathrm{0}} \left(−\mathrm{1}\right)^{\mathrm{10}} \mathrm{1}^{\mathrm{2}} =\frac{\mathrm{12}!}{\mathrm{10}!\mathrm{2}!}=\mathrm{66}\:\checkmark \\ $$
Commented by mr W last updated on 28/Feb/24
