**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.

my friend had this answer for a) 4+11-1 = 14

11

is that right?

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Not sure I understand what you mean by that answer… I seem to get “100 different combos” as stated.

My daughter and I have been terribly sick with the flu, which is why there has been a delay in my posts/comments… Sorry

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ooh ok hope yall both feel better!!

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