Reach X: Find the Path Without Going Through Black Spaces

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To reach the square marked with X while avoiding black spaces, there are two initial paths: moving left then right or right then left to the square directly beneath. After this, there is only one path to the X from the subsequent white square. The only viable route requires passing through a specific white square in the fifth row from the bottom. Thus, the total number of distinct paths to X is three.
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How many different ways can I get to the square marked with X if I can't go through the black spaces?

http://i.imgur.com/HCSoD.png

This is the original image. Essentially, how can I get to X abiding by checker rules (a piece can only move diagonally forwards by one square at a time by only moving on the white squares. How many paths are there in total?


I counted two, because in order to get to the square directly beneath you, you must go either left then right or right then left. After that there is only one path to the x.
There is no way to go around the black squares to the left.

This is correct, right?
 
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Hi SeththeBaller! :wink:
SeththeBaller said:
… Essentially, how can I get to X abiding by checker rules (a piece can only move diagonally forwards by one square at a time by only moving on the white squares. How many paths are there in total?

HCSoD.png


I counted two, because in order to get to the square directly beneath you, you must go either left then right or right then left. After that there is only one path to the x.
There is no way to go around the black squares to the left.

This is correct, right?

Yep, looks like it! :smile:
 


Me, too. All (i.e., both) paths have to go through the white square on the right side, in the 5th row up from the lower left corner. From that square there is only one path to the X.
 


Thanks guys!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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