Answer
$\dfrac{22!}{14!8!}x^{16}y^{14}$
Work Step by Step
Our aim is to find the terms containing $y^{14}$ for $(x^2+y)^{32}$.
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, $(x^2+y)^{32}=\displaystyle \binom{22}{14}(x^2)^{22-14}(y)^{14}$
This implies,
$=\dfrac{22!}{14!(22-14)!}(x)^{16}y^{14}$
$=\dfrac{22!}{14!8!}x^{16}y^{14}$
Thus, the terms containing $y^{14}$ are $\dfrac{22!}{14!8!}x^{16}y^{14}$
