[asy]
label("$1$",(0,0),S);
label("$1$",(-1,-1),S);
label("$1$",(-2,-2),S);
label("$1$",(-3,-3),S);
label("$1$",(-4,-4),S);
label("$1$",(1,-1),S);
label("$1$",(2,-2),S);
label("$1$",(3,-3),S);
label("$1$",(4,-4),S);
label("$2$",(0,-2),S);
label("$3$",(-1,-3),S);
label("$3$",(1,-3),S);
label("$4$",(-2,-4),S);
label("$4$",(2,-4),S);
label("$6$",(0,-4),S);
label("etc.",(0,-5),S);
//Credit to chezbgone2 for the diagram[/asy]

Pascal's triangle is an array of positive integers(See figure), in which the first row is $1$, the second row is two $1$'s, each row begins and ends with $1$, and the $k^\text{th}$ number in any row when it is not $1$, is the sum of the $k^\text{th}$ and $(k-1)^\text{th}$ numbers in the immediately preceding row. The quotient of the number of numbers in the first $n$ rows which are not $1$'s and the number of $1$'s is

$\textbf{(A) }\dfrac{n^2-n}{2n-1}\qquad\textbf{(B) }\dfrac{n^2-n}{4n-2}\qquad\textbf{(C) }\dfrac{n^2-2n}{2n-1}\qquad\textbf{(D) }\dfrac{n^2-3n+2}{4n-2}\qquad \textbf{(E) }\text{None of these}$
1. **Determine the number of 1's in the first \( n \) rows:**
   - The first row has one 1.
   - Each subsequent row has two 1's (one at the beginning and one at the end).
   - Therefore, the total number of 1's in the first \( n \) rows is:
     \[
     1 + 2(n-1) = 1 + 2n - 2 = 2n - 1
     \]

2. **Determine the total number of elements in the first \( n \) rows:**
   - The number of elements in the \( k \)-th row is \( k \).
   - The total number of elements in the first \( n \) rows is the sum of the first \( n \) natural numbers, which is the \( n \)-th triangular number:
     \[
     \sum_{k=1}^n k = \frac{n(n+1)}{2}
     \]

3. **Determine the number of elements that are not 1's:**
   - Subtract the number of 1's from the total number of elements:
     \[
     \frac{n(n+1)}{2} - (2n - 1)
     \]
   - Simplify the expression:
     \[
     \frac{n(n+1)}{2} - 2n + 1 = \frac{n^2 + n - 4n + 2}{2} = \frac{n^2 - 3n + 2}{2}
     \]

4. **Calculate the quotient of the number of non-1 elements and the number of 1's:**
   - The quotient is:
     \[
     \frac{\frac{n^2 - 3n + 2}{2}}{2n - 1}
     \]
   - Simplify the expression:
     \[
     \frac{n^2 - 3n + 2}{2(2n - 1)} = \frac{n^2 - 3n + 2}{4n - 2}
     \]

The final answer is \(\boxed{\frac{n^2 - 3n + 2}{4n - 2}}\)