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What is the coefficient of $x^{3} y^{4}$ in $\left(2 x+3 y^{2}\right)^{5}$ ?
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Verified Answer
The correct answer is:
720
$\mathrm{T}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}(2 \mathrm{x})^{\mathrm{r}-1}\left(3 \mathrm{y}^{2}\right)^{\mathrm{n}-\mathrm{r}+1}$
$\mathrm{~T}_{4}=={ }^{5} \mathrm{C}_{3}(2 \mathrm{x})^{3}\left(3 \mathrm{y}^{2}\right)^{2}$
$=\frac{5 !}{3 ! 2 !} 2^{3} \cdot \mathrm{x}^{3} \cdot 9 \mathrm{y}^{4}=\frac{5.4}{2.1} \times 8 \times 9 \times \mathrm{x}^{3} \mathrm{y}^{4}=720 \mathrm{x}^{3} \mathrm{y}^{4}$
$\therefore$ Coefficient of $x^{3} y^{4}=720$
$\mathrm{~T}_{4}=={ }^{5} \mathrm{C}_{3}(2 \mathrm{x})^{3}\left(3 \mathrm{y}^{2}\right)^{2}$
$=\frac{5 !}{3 ! 2 !} 2^{3} \cdot \mathrm{x}^{3} \cdot 9 \mathrm{y}^{4}=\frac{5.4}{2.1} \times 8 \times 9 \times \mathrm{x}^{3} \mathrm{y}^{4}=720 \mathrm{x}^{3} \mathrm{y}^{4}$
$\therefore$ Coefficient of $x^{3} y^{4}=720$
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