Problem: Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$ , distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$ , and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$ . Additionally, when segments $\overline{A_iB_j}$ are drawn for all $i=1,2,3,\ldots, m$ and $j=1,\allowbreak 2,\allowbreak 3, \ldots, \allowbreak n$ , no point strictly between $\ell_A$ and $\ell_B$ lies on more than 1 of the segments. Find the number of bounded regions into which this figure divides the plane when $m=7$ and $n=5$ . The figure shows that there are 8 regions when $m=3$ and $n=2$ [asy] import geometry; size(10cm); draw((-2,0)--(13,0)); draw((0,4)--(10,4)); label("$\ell_A$",(-2,0),W); label("$\ell_B$",(0,4),W); point A1=(0,0),A2=(5,0),A3=(11,0),B1=(2,4),B2=(8,4),I1=extension(B1,A2,A1,B2),I2=extension(B1,A3,A1,B2),I3=extension(B1,A3,A2,B2); draw(B1--A1--B2); draw(B1--A2--B2); draw(B1--A3--B2); label("$A_1$",A1,S); label("$A_2$",A2,S); label("$A_3$",A3,S); label("$B_1$",B1,N); label("$B_2$",B2,N); label("1",centroid(A1,B1,I1)); label("2",centroid(B1,I1,I3)); label("3",centroid(B1,B2,I3)); label("4",centroid(A1,A2,I1)); label("5",(A2+I1+I2+I3)/4); label("6",centroid(B2,I2,I3)); label("7",centroid(A2,A3,I2)); label("8",centroid(A3,B2,I2)); dot(A1); dot(A2); dot(A3); dot(B1); dot(B2); [/asy]

Solution: When a new point is added to a line, the number of newly bounded regions it creates with each line segment will be one more than the number of intersection points the line makes with other lines.
Case 1: If a new point $P$ is added to the right on a line when both lines have an equal amount of points.
WLOG, let the point be on line $\ell_A$ . We consider the complement, where new lines don't intersect other line segments. Simply observing, we see that the only line segments that don't intersect with the new lines are lines attached to some point that a new line does not pass through. If we look at a series of points on line $\ell_B$ from left to right and a line connects $P$ to an arbitrary point, then the lines formed with that point and with remaining points on the left of that point never intersect with the line with $P$ . Let there be $s$ points on lines $\ell_A$ and $\ell_B$ before $P$ was added. For each of the $s$ points on $\ell_B$ , we subtract the total number of lines formed, which is $s^2$ , not counting $P$ . Considering all possible points on $\ell_B$ , we get $(s^2-s)+(s^2-2s)\cdots(s^2-s^2)$ total intersections. However, for each of the lines, there is one more bounded region than number of intersections, so we add $s$ . Simplifying, we get $s^3-s\sum_{i=1}^{s}{i}+s\Longrightarrow s(s^2-\sum_{i=1}^{s}{i}+1)$ . Note that this is only a recursion formula to find the number of new regions added for a new point $P$ added to $\ell_A$
Case 2:  If a new point $P$ is added to the right of a line that has one less point than the other line.
Continuing on case one, let this point $P$ be on line $\ell_B$ . With similar reasoning, we see that the idea remains the same, except $s+1$ lines are formed with $P$ instead of just $s$ lines. Once again, each line from $P$ to a point on line $\ell_A$ creates $s$ non-intersecting lines for that point and each point to its left. Subtracting from $s(s+1)$ lines and considering all possible lines created by $P$ , we get $(s(s+1)-s)+(s(s+1)-2s)\cdots(s(s+1)-s(s+1)$ intersections. However, the number of newly bounded regions is the number of intersections plus the number of points on line $\ell_A$ . Simplying, we get $s(s+1)^2-s\sum_{i=1}^{s+1}{i}+(s+1)$ newly bounded regions.
For the base case $s=2$ for both lines, there are $4$ bounded regions. Next, we plug in $s=2,3,4$ for both formulas and plug $s=5$ for the first formula to find the number of regions when $m=6$ and $n=5$ . Notice that adding a final point on $\ell_A$ is a variation of our Case 1. The only difference is for each of the $s$ lines formed by $P$ , there are $s+1$ points that can form a non-intersecting line. Therefore, we are subtracting a factor of $s+1$ lines instead of $s$ lines from a total of $s(s+1)$ lines. However, the number of lines formed by $P$ remains the same so we still add $s$ at the end when considering intersection points. Thus, the recursive equation becomes $(s(s+1)-(s+1))+(s(s+1)-2(s+1))\cdots(s(s+1)-s(s+1))+s\Longrightarrow s^2(s+1)-(s+1)\sum_{i=1}^{s}{i}+s$ . Plugging $s=5$ into this formula and adding the values we obtained from the other formulas, the final answer is $4+4+9+12+22+28+45+55+65=\boxed{244}$

Answer:  244