Q:  Here are some counting problems:

(a) How may different positive integer answers are to x1 + x2 + x3 + x4 = 11?
(b) How may different positive integer answers are to x1 + x2 + x3 <= 14?

A:  I have been thinking about this problem for a while and do not yet know if there is a better way to start this other than just straight out counting:

(a)  x1 + x2 + x3 + x4 = 11

So, one answer could be (1, 1, 1, 8), for example.  Another answer could be (1, 1, 8, 1)… However, this answer has the same digits of the first answer… I will worry about the different combinations later!  So:

(1, 1, 1, 8 )

(1, 1, 2, 7)

(1, 1, 3, 6)

(1, 1, 4, 5)

(1, 2, 2, 6)

(1, 2, 3, 5)

(1, 2, 4, 4)

(1, 3, 3, 4)

OK.  So, I believe that is all the permutations.  Did I miss any?  But now I need to count how many combinations there are of each:

(1, 1, 1, 8 ) = 4!/3! = 4 combos

(1, 1, 2, 7) = 4!/2! = 12 combos

(1, 1, 3, 6) = 4!/2! = 12 combos

(1, 1, 4, 5) = 4!/2! = 12 combos

(1, 2, 2, 6) = 4!/2! = 12 combos

(1, 2, 3, 5) = 4! = 24 combos

(1, 2, 4, 4) = 4!/2! = 12 combos

(1, 3, 3, 4) = 4!/2! = 12 combos

Add up all the red to get:  100 total combinations

(b)  x1 + x2 + x3 <= 14 (that sign is “less than or equal to”)

Again, the first method that comes to my mind is to systematically count (and then worry about combinations *different orders* later):

(x1, x2, x3) =

(1, 1, 1)   (1, 1, 2) (1, 1, 3) …. (1, 1, 12)

(1, 2, 2)   (1, 2, 3) (1, 2, 4) …. (1, 2, 11)

(1, 3, 3)   (1, 3, 4) (1, 3, 5) …. (1, 3, 10)

(1, 4, 4)   (1, 4, 5) (1, 4, 6) …. (1, 4, 9)

(1, 5, 5)   (1, 5, 6) (1, 5, 7) …. (1, 5, 8 )

…. Screaming baby! Must continue soon!….  There must be a better way thant this…. but, it will work.