Question 1210219: Find the number of paths from A to B in the grid below, so that
* Each step is down or to the right.
* The path cannot pass through any point more than once.
An example path is shown.
The grid is 3 by 3, with A at the upper-left, and B at the lower-right.
Found 2 solutions by CPhill, greenestamps:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's analyze this problem carefully. We are looking for paths from A(0,0) to B(3,3) using only steps down (D) or to the right (R), with the added constraint that the path cannot pass through any point more than once (which is automatically satisfied if we only move down or right in a grid).
In a $3 \times 3$ grid, to go from A to B, we need to make a total of 3 steps to the right and 3 steps down, in any order. The total number of steps is $3 + 3 = 6$.
If there were no restrictions other than moving down or right, the number of paths would be given by the number of ways to choose 3 right steps (or 3 down steps) out of the total 6 steps, which is $\binom{6}{3} = \frac{6!}{3!3!} = \frac{720}{6 \times 6} = 20$.
However, the image shows a grid, and it seems the question implies a specific grid configuration where paths might be obstructed or the movement is constrained within the lines of the grid. Assuming the standard interpretation of a grid path from the top-left to the bottom-right using only down and right steps, the constraint of not passing through any point more than once is inherently satisfied.
Let's re-read the question to ensure no details were missed. "Find the number of paths from A to B in the grid below..." The "grid below" is described as 3 by 3. If it's a standard grid, the number of steps required is consistent.
Consider the movement on the grid lines. To go from the top-left intersection to the bottom-right intersection of a $3 \times 3$ grid (meaning 3 units down and 3 units to the right), any path will consist of a sequence of 3 'R' moves and 3 'D' moves. The number of such sequences is $\binom{6}{3} = 20$.
The example path shown in the image confirms this type of movement. The path moves along the grid lines, either down or to the right, at each step.
The constraint "The path cannot pass through any point more than once" is redundant for paths restricted to down and right movements in a grid from the top-left to the bottom-right. Such a path will always be monotonically increasing in both the x and y coordinates, thus never revisiting a point.
Therefore, the problem reduces to finding the number of sequences of 3 'R' moves and 3 'D' moves.
Number of paths = $\binom{\text{total steps}}{\text{number of right steps}} = \binom{6}{3} = 20$.
Alternatively,
Number of paths = $\binom{\text{total steps}}{\text{number of down steps}} = \binom{6}{3} = 20$.
Final Answer: The final answer is $\boxed{20}$
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
If every move is either down or to the right, then obviously the path can't pass through any point more than once....
So to get from A to B we need to move, in some order, 3 steps to the right and 3 units down.
Represent a move to the right as "R" and a move down as "D". Then each distinct path is represented by a string of 3 "R"s and 3 "D"s. By a common counting principle, the number of ways of arranging the symbols "RRRDDD" is
ANSWER: 20
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