Number of Paths on 5x5 Chessboard: Solve the Puzzle!

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To solve the problem of finding the number of paths on a 5x5 chessboard from the lower left corner to the upper right corner, one can only move right or up. The correct calculation involves making 4 moves up and 4 moves right, totaling 8 moves. The number of distinct paths can be determined using the formula for combinations, specifically \(\binom{8}{4} = \frac{8!}{4!4!}\). For a generic nxn board, the formula is \(\binom{2n-2}{n-1} = \frac{(2n-2)!}{(n-1)!(n-1)!}\). This discussion clarifies the correct number of moves and the applicable formula for calculating the paths.
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I know there is a formula for how many of a certain thing like squares on a 8x8 chessboard but I haven't come across anything on pathways .Is there one for this type of problem?Any help would be appreciated.
Consider a 5-by-5 chessboard. You want to move a nickel from the lower left corner to the upper right corner. You are only allowed to move the nickel one square at a time, and each move must be either to the right or up. How many different paths are possible?
Thanks
 
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For an 8x8 board...
If you can only move up or right, then that simplifies things considerably. Note that no matter which path you choose, you must move right 7 times and up 7 times, so we have 14 "moves" in total, no matter which path you choose. Basically, we have 14 slots and want to know how many ways there are to insert 7 ups and 7 rights into those slots. Really, we just want to know the number of ways to arrange 7 objects in 15 slots, since we then get the other 7 for free (just use the empty slots). The solution is then...
\displaystyle \binom{14}{7} = \frac{14!}{7!7!}

For a generic nxn board it would be...

\displaystyle \binom{2n-2}{n-1} = \frac{(2n-2)!}{(n-1)!(n-1)!}
 
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never mind I didnt see the edit.
 
westgrant88 said:
Just wondering how you get 7 moves right and 7 up. Icount 4 and 4. How am I missing 3?

That's my bad; I did it for an 8x8. For a 5x5, you're right, it would be 4 moves up and 4 moves right.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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