Question

A man starts at the origin in the coordinate plane. He can crawl one unit length to the right or one unit up. He can only crawl to the right, or up during his journey; not down, to the left, or off the coordinate grid lines! He is trying to get to the coordinate (4,3). How many different paths can he take to reach the point (4,3)? A man starts at the origin in the coordinate plane. He can crawl one unit length to the right or one unit up. He can only crawl to the right, or up during his journey; not down, to the left, or off the coordinate grid lines! He is trying to get to the coordinate (4,3). How many different paths can he take to reach the point (4,3)? @Mathematics

Answer 1

any help would be appreciated

Answer 2

I am getting an answer of 31.... I counted the numer of squares (12) , then multiplied by the number of surfaces (4) and got 48... I then subtracted the number of common surfaces... is there a formua one could use to come up with the answer?

Answer 3

Why don't you try using a tree, like,

Answer 4

Each branch should have 7 steps, 3 ups and 4 lefts.

Answer 5

looking at your tree now....

Answer 6

does that end up with 31 using the tree?

Answer 7

I have to go unfortunately. It's a combinatorial problem.

Answer 8

If you haven't sorted it by the time I get back, I'll see what I can do.

Answer 9

thanks.... I will review "combinatorial" and see if it helps.
thanks for taking the time.

Answer 10

I will make a comment if I figure it out... thanks again.

Answer 11

you're welcome. good luck.

Answer 12

Did you get 35? I tried accessing the site on my phone after I left but couldn't log in. Like I said, it's a combinatorial problem. You have 7 places to fill with 3 'ups' (U) and 4 'lefts' (L). If each U and L were distinct (e.g. if one left differed from another, and the same for ups), the total number of permutations would be 7!, but since the U's and L's are *not* distinct, we've over-counted. There are 3! permutations for the Us and 4! permutations for the Ls. So the total number of distinct lists is \[\frac{7!}{3!4!}=35\]I don't know what level of mathematics you're studying, which is why I suggested the tree. Let me know if you want further clarification.