Show the number of arrangements that give an overall length of L = 2md

[Another]

Homework Statement

Show that the number of arrangements s that give an overall length
of L = 2md is given by g(N,m) = 2N!/[(N/2+m)!(N/2-m)!]

Relevant Equations

N = N_+ + N_-
L = 2md = (N_+ - N_-)d then 2m = N_+ - N_- , m is positive

I know $L = 2md = (N_+ - N_-)d$ then $2m = N_+ - N_-$
So I can write $N_+$ and $N_-$ in term N and m
I don't understand the factor 2 multiplying in front of N!/[(N_+)!(N_-)!]
How does multiplication by the number "2" give a physical meaning?

 

[PeroK]

I know $L = 2md = (N_+ - N_-)d$ then $2m = N_+ - N_-$
So I can write $N_+$ and $N_-$ in term N and m
I don't understand the factor 2 multiplying in front of N!/[(N_+)!(N_-)!]
How does multiplication by the number "2" give a physical meaning?

Suppose $N = 4$ and $m = 1$: you want a length of $L = 2d$. You need $3$ links in one direction and $1$ in the other. You can do that with $3$ links in the plus direction and $1$ link in the negative direction or vice versa. We can count the number of ways to get $3$ pluses and $1$ minus, then double it.

 

[anuttarasammyak]

2md and -2md are regarded same so doubled.