$11 \cdot 7$ polynomial $(x+y)^{9}$ expanded in descending powers of $x$. When $x=p$ and $y=q$, the second and third terms have equal values, where $p$ and $q$ are positive numbers and their sum is $1$. The value of $p$ is
(A) $\frac{1}{5}$.
(B) $\frac{4}{5}$.
(C) $\frac{1}{4}$.
(D) $\frac{3}{4}$.
(E) $\frac{8}{9}$.
(26th American High School Mathematics Examination, 1975)
[Solution] From $(x+y)^{9}=x^{9}+9 x^{8} y+36 x^{7} y^{2}+\cdots+y^{9}$; and from the given condition $9 p^{8} q=36 p^{7} q^{2}$, we have $\quad p=4 q$.

From $p+q=1$, we get $p=\frac{4}{5}$. Therefore, the answer is $(B)$.