In that case,
The ending point of the path Jeff made depends on all the moves Jeff made, not the order. Consider a random 10-step path Jeff could take. NESWNNESEW (N for North, S for south, W for west, E for East. Reduce our path by cancelling each E-W and each N-S pair, which will not change our endpoint. So our example reduces to NE. Notice that our reduced path must have an even number of moves, and it is thus possible to end at any point (x,y) where x+y is even and |x| + |y| is no greater than 10.'
If |x| + |y| = 0, there is one possibility for (x,y).
If |x| + |y| <= (less than equal to) 2, there are 9 possible paths ((0,0) (-+1,-+1) (0, +-2) (+-2,0)); these solutions will form a 3x3 diagonal grid
The solutions with |x| + |y| <= 10 will form an 11x11 grid, so there are 11^2=121 ways
Sorry if I didn't explain this very well, I'm in a bit of a rush.