MHB How Many Final Points Can Be Reached in a Ten-Step Lattice Walk?

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In a ten-step lattice walk starting from the origin, the object can move in four possible directions: right, left, up, or down. Each step allows for a combination of movements, leading to various potential final positions. The maximum distance from the origin after ten steps is five units in any direction, as the object can only move a total of ten units. The final points are determined by the combinations of movements, resulting in a specific number of reachable lattice points. The discussion emphasizes the combinatorial nature of the problem and encourages participants to explore the solution further.
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Here is this week's POTW:

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An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at the origin and takes a ten-step path, how many different points could be the final point?

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No one answered last week's POTW. (Sadface) However, you can find the suggested solution as follows:
Each step changes either the $x$-coordinate or the $y$-coordinate of the object by 1. Thus if the object’s final point is $(a,\,b)$, then $a+b$ is even and $|a|+|b|\le10$. Conversely, suppose that $(a,\,b)$ is a lattice point with $|a|+|b|= 2k\le10$. One ten-step path that ends at $(a,\,b)$ begins with $|a|$ horizontal steps, to the right if $a\ge 0$ and to the left if $a <0$. It continues with $|b|$ vertical steps, up if $b\ge 0$ and down if $b <0$. It has then reached $(a,\,b)$ in 2k steps, so it can finish with 5−k steps up and 5−k steps down. Thus the possible final points are the lattice points that have even coordinate sums and lie on or inside the squarewith vertices $(\pm10,0)$ and $(0,\pm 10)$. There are 11 such points on each of the 11 lines $x+y= 2k$, $−5\le k \le 5$, for a total of 121 different points.
 

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