## Precalculus (6th Edition) Blitzer

$319,770x^{16}y^{14}$
Calculate the terms containing $y^{14}$ for $(x^2+y)^{22}$ 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,$(x^2+y)^{32}=\displaystyle \binom{22}{14}(x^2)^{22-14}(y)^{14}$ or,$=\dfrac{22!}{14!(22-14)!}(x)^{16}y^{14}$ or,$=\dfrac{22!}{14!8!}x^{16}y^{14}$ or, $=319,770x^{16}y^{14}$ Hence, Terms containing for $y^{14}$ are: $319,770x^{16}y^{14}$