Let f(n) = number of ways of achieving total n.

These can be achieved be adding a 1 to the left of all ways of achieving (n-1), or by adding a 2 to the left of all ways of achieving (n-2), or by adding a 3 to the left of all ways of achieving (n-3), or by adding a 4 to the left of all ways of achieving (n-4). 

In other words, f(n) = f(n-4) + f(n-3) + f(n-2) + f(n-1).  It's an extended fibonacci.

f(0) = 1

f(1) = 1

f(2) = 2

f(3) = 4

f(4) = 8 (ie, 1+1+2+4)

f(5) = 15 (ie, 1+2+4+8)

f(6) = 29 (ie, 2+4+8+15)

f(7) = 56

f(8) = 108

f(9) = 208

f(10) = 401

f(11) = 773

f(12) = 1490