Quick No Programming Solution (based on combinatorics)
I take it “no backtracking” means we always either increase x or increase y.
If so, we know that in total we will have 40 steps to reach the finish — 20 increases in x, 20 increases in y.
The only question is which of the 40 are the 20 increases in x. The problem amounts to: how many different ways can you choose 20 elements out of a set of 40 elements. (The elements are: step 1, step 2, etc. and we’re choosing, say, the ones that are increases in x).
There’s a formula for this: it’s the binomial coefficient with 40 on top and 20 on the bottom. The formula is 40!/((20!)(40-20)!), in other words 40!/(20!)^2. Here ! represents factorial. (e.g., 5! = 5*4*3*2*1)
Canceling out one of the 20! and part of the 40!, this becomes: (40*39*38*37*36*35*34*33*32*31*30*29*28*27*26*25*24*23*22*21)/(20*19*18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1). The problem is thus reduced to simple arithmatic. The answer is 137,846,528,820.
For comparison, note that (4*3)/(2*1) gives the answer from their example, 6.