$11 \cdot 9$ As shown, Pascal's (Yang Hui's) Triangle is an array of positive integers, where the first row is 1, the second row is two 1s, each row starts and ends with 1, and in any row, the $k$-th number (when it is not 1) is the sum of the $k$-th and $(k-1)$-th numbers in the previous row. In the first $n$ rows, the ratio of the number of numbers that are not 1 to the number of 1s is
$$
\begin{array}{ccccc} 
& & 1 & & \\
& 1 & 1 & & \\
& 1 & 2 & 1 & \\
& 1 & 3 & 3 & 1 \\
1 & 4 & 6 & 4 & 1
\end{array}
$$
(A) $\frac{n^{2}-n}{2 n-1}$.
(B) $\frac{n^{2}-n}{4 n-2}$.
(C) $\frac{n^{2}-2 n}{2 n-1}$.
(D) $\frac{n^{2}-3 n+2}{4 n-2}$.
(E) None of these.
(22nd American High School Mathematics Examination, 1971)
[Solution] In the first $n$ rows, the number of 1's is $(2 n-1)$, and the number of other numbers is
$$
\frac{1}{2}(n-2)(n-1)=\frac{1}{2}\left(n^{2}-3 n+2\right) \text { . }
$$

The quotient is
$$
\frac{n^{2}-3 n+2}{4 n-2} \text {. }
$$

Therefore, the answer is $(D)$.