Answer
$=x^{4}+8x^{3}y+24x^{2}y^{2}+32xy^{3}+16y^{4}$
Work Step by Step
$\left( x+2y\right) ^{4}= \binom{4}{0} \times x^{4} \times(2y)^{0}+\binom{4}{1} \times x^{3} \times(2y)^{1}+\binom{4}{2} \times x^{2} \times(2y)^{2}+\binom{4}{3} \times x^{1} \times(2y)^{3}+\binom{4}{4} \times x^{0} \times(2y)^{4}$
$\binom{4}{0}=\binom{4}{4}=\dfrac {4!}{0!\times 4!}=1$
$\binom{4}{1}=\binom{4}{3}=\dfrac {4!}{3!}=4$
$\binom{4}{1} =\dfrac {4!}{2!\times \left( 4-2\right) !}=\dfrac {2!\times 3\times 4}{2!\times 2!}=6$
So we get
$\left( x+2y\right) ^{4}=x^{4}+4x^{3}\times \left( 2y\right) +6x^{2}\times \left( 2y\right) ^{2}+4\times \left( 2y\right) ^{3}+\left( 2y\right) ^{4}=x^{4}+8x^{3}y+24x^{2}y^{2}+32xy^{3}+16y^{4} $
