Answer
$13440x^{4}y^{6}$
Work Step by Step
Our aim is to find the terms containing $y^{6}$ for $(x+2y)^{10}$.
General formula:$(p+q)^n=\displaystyle \binom{n}{k}p^{n-k}q^k$
and $\displaystyle \binom{n}{k}=\dfrac{n!}{k!(n-k)!}$
Now, $(x+2y)^{10}=\displaystyle \binom{10}{6}(x)^{10-6}(2y)^{6}$
This implies,
$=\dfrac{10!}{6!(10-6)!}(64)(x)^{4}(y)^{6}$
$=\dfrac{10 \times 9 \times 8 \times 7 \times 6!}{(6!)(4 \times 3 \times 2 \times 1)}x^{16}y^{14}$
Thus, the terms containing $y^{6}$ are $13440x^{4}y^{6}$
