## mathmath333 one year ago Counting Question

1. mathmath333

|dw:1440284134239:dw|

2. mathmath333

\large \color{black}{\begin{align} & \normalsize \text{The Red and Green lines are perpendicular to each other}\hspace{.33em}\\~\\ & \normalsize \text{Find the number of shortest paths to go from A to B.}\hspace{.33em}\\~\\ & \normalsize \text{Find the total number of paths to go from A to B.}\hspace{.33em}\\~\\ & \normalsize \text{(Assume the points are equidistant)}\hspace{.33em}\\~\\ \end{align}}

3. Miracrown

You're welcom

4. Loser66

I'm on iPad , cannot draw out the paths, b to me , there are 2 shortest ways , and their weight is 6

5. zzr0ck3r

I will go look at my graph theory book and I am sure I will find something.

6. mathmath333

are u sure its 6

7. Loser66

Go, diagonal fromA, 3 times, then go straight to B 3 more

8. zzr0ck3r

there are algorithms for this, that would be a pain to teach on here. google Dijkstra's algorithm

9. mathmate

Assume each of the 6 red spaces have the same distance as the 3 green spaces, Number of paths = 9!/(6!3!)=84 ways

10. zzr0ck3r

diagonal?

11. mathmate

along the lines, if the question was for me.

12. mathmath333

Djikistra algorithm finds shortest distance, where as i need the number of shortest paths

13. Loser66

Go to the right of A 1, then, diagonal, then right,then,diagonal,repeat 1 more time to get B

14. Loser66

Use adjacent matrix to find them out. It works also

15. Loser66

Why not diagonal? Since it is = right+ vertical up.?

16. mathmath333

answer is 210 for 1st one

17. mathmath333

the path has to along the lines can't jump diagonally

18. Loser66

Ohoh, so I'm wrong, hehe.

19. mathmath333

|dw:1440285501178:dw| an example

20. zzr0ck3r

Yes that is right, there is some algorithm that will tell you.

21. mathmate

It's like a permutation with 9 objects, 3 red and 6 green. RGRGRGGGG RRRGGGGGG .... Using the multinomial theorem, the number of ways would be 9!/(6!3!)

22. mathmath333

thanks @mathmate ur anwers was correct

23. mathmate

You're welcome! :)

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