19. As shown in Figure 4, in rectangle $A B C D$, $A B=6, B C$ $=3$, point $E$ is on side $B C$, point $F$ is on side $C E$, $B E=$ $E F=F C$. Connecting $A E$ and $A F$, they intersect $B D$ at points $P$ and $Q$ respectively. If $B P: P Q: Q D=r: s: t\left(r, s, t \in \mathbf{Z}_{+},(r, s, t)=1\right)$, then $r+s+t=(\quad)$.
(A) 7
(B) 9
(C) 12
(D) 15
(E) 20
19. E.

From $A D / / B C \Rightarrow \triangle A P D \backsim \triangle E P B$
$$
\Rightarrow \frac{r}{s+t}=\frac{B E}{A D}=\frac{1}{3} \text {. }
$$

Similarly, $\frac{r+s}{t}=\frac{B F}{A D}=\frac{2}{3}$.
$$
\begin{array}{l}
\text { Then }\left\{\begin{array}{l}
s+t=3 r, \\
3 r+3 s=2 t
\end{array} \Rightarrow t=4 s \Rightarrow 5 s=3 r\right. \\
\Rightarrow r: s: t=5: 3: 12 \\
\Rightarrow r+s+t=5+3+12=20 .
\end{array}
$$
20. B.

Assume each term in the expansion contains $a^{x} b^{y} c^{z} d^{x} 1^{t}$. Then
$$
x+y+z+w+t=N,
$$

where $x, y, z, w \in \mathbf{Z}_{+}, t \in \mathbf{N}$.
Let $x^{\prime}=x-1, y^{\prime}=y-1, z^{\prime}=z-1, w^{\prime}=w-1$.
Thus, $x^{\prime}+y^{\prime}+z^{\prime}+w^{\prime}+t=N-4$.
Note that, the number of non-negative integer solutions to the indeterminate equation
$$
x^{\prime}+y^{\prime}+z^{\prime}+w^{\prime}+t=N-4
$$

is the same as the number of terms in the expansion of the polynomial $(a+b+c+d+1)^{N}$.
Therefore, $\mathrm{C}_{N}^{4}=1001 \Rightarrow N=14$.