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Q.
What is the coefficient of $x^{3} y^{4}$ in $\left(2 x+3 y^{2}\right)^{5} ?$
Binomial Theorem
Solution:
$T _{ r }={ }^{ n } C _{ r -1}(2 x )^{ r -1}\left(3 y ^{2}\right)^{ n - r +1}$
$T _{4}= T _{3+1}={ }^{5} C _{3}(2 x )^{3}\left(3 y ^{2}\right)^{2}$
$=\frac{5 !}{3 ! 2 !} 2^{3} . x^{3} . 9 y^{4}=\frac{5.4}{2.1} \times 8 \times 9 \times x^{3} y^{4}=720 \,x^{3} y^{4}$
$\therefore $ Coefficient of $x^{3} y^{4}=720$