Answer
$13440x^{4}y^{6}$
Work Step by Step
Calculate the terms containing $y^{6}$ for $(x+2y)^{10}$ by using General formula such as:$(m+n)^r=\displaystyle \binom{r}{k}m^{r-k}n^k$
and $\displaystyle \binom{r}{k}=\dfrac{r!}{k!(r-k)!}$
This implies,$(x+2y)^{10}=\displaystyle \binom{10}{6}x^{(10-6)}(2y)^{6}$
or,$=(2)^6[\dfrac{10!}{6!(10-6)!}]x^{4}(y)^{6}$
$=[\dfrac{10 \times 9 \times 8 \times 7 \times 6!}{6!(4 \times 3 \times 2 \times 1)}](x)^{16}y^{14}$
Hence, Terms containing for $y^{6}$ are: $13440x^{4}y^{6}$
