How many paths are there from A to B along vertical and horizontal lines?

Question

There are 4 horizontal lines and 5 vertical lines. A is in the top left corner and B is at the lower right. The problem is in the permutation/combination chapter, and may or may not deal with the binomial theorem. I am completely clueless on how to solve this.

Answer 1

I believe this is a Pascal's Triangle problem
....................1
....................1..1
..................1...2...1
................1...3...3...1
and it goes on.

Put these numbers on the vertices of the squares where A is the first 1 and continue with the pattern....I got 35.

Answer 2

Drawing a tree of possibilities is one way to go.
On the top horizontal line, name the points of intersection with the five vertical lines A, C, D, E, and F. On the second horizontal line name the corresponding points G,H,I,J, and K. On the 3rd line name the points L,M,N,O and P. On the bottom line name the points Q,R,S,T,B.
Now starting at point A, there are two possible paths. You could go to Point C or Point G. So you begin your tree with Point A and have two lines coming out of it at about 45 degree angles, and ending at a Point labeled C and a Point labled G.

Now from point G you can go to either point H or point L. And from Point C you can go to either point D or point H. So now you expand your tree so that 2 lines emanate from point C ending at Points D and H and 2 lines emanating from point G ending in points H and L.

Continue in this way and you will see a pattern build up that you will be able to formulate. Just make sure you don't allow any moves that retrace your actions For example, don't go from A to B and the back to A and try to say this is a different path.

Try this approach. I think you'll like it.